Approximately what percent of normally distributed data values lie within 2 standard deviation to either side of the mean?

According to a​ publication, 12.3​% of 18 to 25 dash year dash olds were users of marijuana in 2000. A recent poll of 1293 randomly selected 18 to 25 dash year dash olds revealed that 176 currently use marijuana. At the 5​% significance​ level, do the data provide sufficient evidence to conclude that the percentage of 18 to 25 dash year dash olds who currently use marijuana has changed from the 2000 percentage of 12.3​%? Use the​ one-proportion z-test to perform the appropriate hypothesis​ test, after checking the conditions for the procedure. What are the hypotheses for the​ one-proportion z-test? Upper H 0​: pequals nothing​; Upper H Subscript a​: p ▼ greater than not equals less than nothing ​(Type integers or​ decimals.)

In a​ study, researchers wanted to measure the effect of alcohol on the hippocampal​ region, the portion of the brain responsible for​ long-term memory​ storage, in adolescents. The researchers randomly selected 22 adolescents with alcohol use disorders to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal volume of 9.02cm3. An analysis of the sample data revealed that the hippocampal volume is approximately normal with

x overbarxequals=8.06cm3 and s=0.7cm3. Conduct the appropriate test at the α=0.01 level of significance.

(a) State the null and alternative hypothese

(b) Identify the t- statistic and p value

(c) Conclusion: (reject/Fail to reject) __ the null hypothesis. There (is/is not) __ sufficent evidence to claim that the mean hippocampal volume is (equal to/less than/greater than) __ __ cm3

1. An experiment was run to examine the amount of time it takes to boil a given amount of water on the four different burners of her stove, and with 0, 2, 4, or 6 teaspoons of water. The numbers in parentheses are run order. The results of the design are given below. Use a=0.05 unless otherwise specified

1. Analyze the full model and check for significance
2. Reduce your model
3. Check the adequacy of this model
4. Determine which settings yield the shortest time

The manager of the Groove Shopping Mall in Windhoek claims that visitors to this mall spend always more than 90 minutes in the mall on any occasion. to test this claim, the chairman of the Namibian chamber of commerce and Industry (NCCI) commissioned a study which found that, from a random sample of 25 recent visitors to this mall, the average visiting time was 95.5 minutes, with a standard deviation of 15 minutes. 2.3.1 Formulate a suitable null and alternative hypothesis for this situation 2.3.2 Compute the sample statistic 2.3.3 formulate the decision rule by using = 10% 2.3.4 What conclusion can the chairman of the Namibian Chamber of Commerce and Industry (NCCI) draw from the Findings?

The Cotton Mill is an upscale chain of women's clothing stores, located in the southwestern United States. Do to recent success, The Cotton Mill's top management is planning to expand by locating new stores in other regions of the country. The director of planning has been asked to study the relationship between yearly sales and the store size. As part of the study, the director selects a sample of 25 stores and determines the size of the store in square feet and the sales for the last year. The sample data follows.

Store size (1000s of square feet)        Sales ( millions of $ )

     3.7                                                9.18

     2.0                                                4.58

     5.0                                                8.22

     0.7                                                1.45

     2.6                                                6.51

     2.9                                                2.82

     5.2                                               10.45

     5.9                                                 9.94

     3.0                                                 4.43

     2.4                                                 4.75

     2.4                                                 7.30

   0.5                                                 3.33

   5.0                                                 6.67

   0.4 0.55

   4.2 7.56

   3.1 2.23

   2.6 4.49

   5.2 9.90

   3.3 8.93

   3.2 7.60

   4.9 3.71

   5.5 5.47

   2.9 8.22

   2.2 7.17

   2.3 4.35

Using store size as the independent variable, run the data using excel and answer the following: 1. Write the regression equation. 2. Interpret the regression constant and regression coefficient. 3. Forecast a value for the dependent variable. 4. Test the significant of the regression coefficient using alpha = .05. 5. Test the overall significant of the regression model. 6. Interpret the coefficient of determination.  

Explain what it means conceptually when we find that a correlation — or any other statistic that is testing a hypothesis — is significant. Refer to type I error, alpha, and confidence level.

A consumer research organization was interested in the influence of type of water on the effectiveness of a detergent. Test batches of washings were run in four randomly chosen machines having a particular type of water - soft, moderate, and hard. All batches had equal numbers of oil-stained rags and after each washing the number of rags stilled stained was determined. The following results were obtained.

                                     Number of Rags with Stains

              Sample

             Observations        Soft      Moderate      Hard

                      1                   0             4              10

                      2                   1             8                5

                      3                   2             3                7

                      4                   1             5               10

Using an alpha level of .01, would you conclude that the type of water influences the effectiveness of the detergent?

Critics of television often refer to the detrimental effects that all the violence shown on television has on children. However, there may be another problem. It may be that watching television also reduces the amount of physical exercise, causing weight gains. A sample of 15 10-year-old children was taken. The number of pounds each child was overweight was recorded (a negative number indicates the child is underweight). In addition, the number of hours of television viewing per week was also recorded. These data are listed here. (also provided in the Excel Spreadsheet) Television 42 34 25 35 37 38 31 33 Overweight 18 6 0 −1 13 14 7 7 Television 19 29 38 28 29 36 18 Overweight −9 8 8 5 3 14 −7 REQUIRED a) Draw the scatter diagram. b) Calculate the sample regression line and describe what the coefficients tell you about the relationship between the two variables.

You are a member of a group of researchers who study the reading ability of children. You would like to know if any of the following factors have an effect on the reading ability of a child: age, memory span, and IQ. You conduct a pilot study on a small group of 20 children. From the initial finding, you will make recommendations to your group on further research.

A. Build three different regression models the using the variables as shown below. Review the results and determine which is the model is the “best”.

1. What is the ANOVA table of the best model?

2. What is the regression equation of the best model?

3. Conduct the test for the significance of the Overall Regression Model for the best model?

4. What is R² of the best model?

5. For the best model, what are the 95% confidence intervals for the estimates of the regression coefficients—the B_i’s?

6. Provide an interpretation of the slopes, b_i’s.

Compare and contrast the deductive and inductive research models.

The following table shows the Myers-Briggs personality preferences for a random sample of 406 people in the listed professions. E refers to extroverted and I refers to introverted.

Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.05 level of significance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: Myers-Briggs preference and profession are not independent
H₁: Myers-Briggs preference and profession are not independent.H₀: Myers-Briggs preference and profession are independent
H₁: Myers-Briggs preference and profession are not independent.    H₀: Myers-Briggs preference and profession are not independent
H₁: Myers-Briggs preference and profession are independent.H₀: Myers-Briggs preference and profession are independent
H₁: Myers-Briggs preference and profession are independent.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

YesNo    

What sampling distribution will you use?

Student's tuniform    chi-squarenormalbinomial

What are the degrees of freedom?


(c) Find or estimate the P-value of the sample test statistic.

p-value > 0.1000.050 < p-value < 0.100    0.025 < p-value < 0.0500.010 < p-value < 0.0250.005 < p-value < 0.010p-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is insufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.

The following table shows age distribution and location of a random sample of 166 buffalo in a national park.

Use a chi-square test to determine if age distribution and location are independent at the 0.05 level of significance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: Age distribution and location are not independent.
H₁: Age distribution and location are not independent.H₀: Age distribution and location are independent.
H₁: Age distribution and location are not independent.    H₀: Age distribution and location are not independent.
H₁: Age distribution and location are independent.H₀: Age distribution and location are independent.
H₁: Age distribution and location are independent.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

YesNo    

What sampling distribution will you use?

Student's tbinomial    normalchi-squareuniform

What are the degrees of freedom?


(c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)

p-value > 0.1000.050 < p-value < 0.100    0.025 < p-value < 0.0500.010 < p-value < 0.0250.005 < p-value < 0.010p-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is sufficient evidence to conclude that age distribution and location are not independent.At the 5% level of significance, there is insufficient evidence to conclude that age distribution and location are not independent.    

The following table shows the Myers-Briggs personality preference and area of study for a random sample of 519 college students. In the table, IN refers to introvert, intuitive; EN refers to extrovert, intuitive; IS refers to introvert, sensing; and ES refers to extrovert, sensing.

Use a chi-square test to determine if Myers-Briggs preference type is independent of area of study at the 0.05 level of significance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: Myers-Briggs type and area of study are independent.
H₁: Myers-Briggs type and area of study are not independent.H₀: Myers-Briggs type and area of study are not independent.
H₁: Myers-Briggs type and area of study are independent.    H₀: Myers-Briggs type and area of study are not independent.
H₁: Myers-Briggs type and area of study are not independent.H₀: Myers-Briggs type and area of study are independent.
H₁: Myers-Briggs type and area of study are independent.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

YesNo    

What sampling distribution will you use?

Student's tnormal    chi-squarebinomialuniform

What are the degrees of freedom?


(c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)

p-value > 0.1000.050 < p-value < 0.100    0.025 < p-value < 0.0500.010 < p-value < 0.0250.005 < p-value < 0.010p-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs type and area of study are not independent.At the 5% level of significance, there is insufficient evidence to conclude that Myers-Briggs type and area of study are not independent.   

Salmon Weights: Assume that the weights of spawning Chinook salmon in the Columbia river are normally distributed. You randomly catch and weigh 17 such salmon. The mean weight from your sample is 22.2 pounds with a standard deviation of 4.4 pounds. You want to construct a 95% confidence interval for the mean weight of all spawning Chinook salmon in the Columbia River.

(a) What is the point estimate for the mean weight of all spawning Chinook salmon in the Columbia River?
pounds

(b) Construct the 95% confidence interval for the mean weight of all spawning Chinook salmon in the Columbia River. Round your answers to 1 decimal place.
< μ <   

(c) Are you 95% confident that the mean weight of all spawning Chinook salmon in the Columbia River is greater than 18 pounds and why?

No, because 18 is above the lower limit of the confidence interval.Yes, because 18 is below the lower limit of the confidence interval.     No, because 18 is below the lower limit of the confidence interval.Yes, because 18 is above the lower limit of the confidence interval.

(d) Recognizing the sample size is less than 30, why could we use the above method to find the confidence interval?

Because the parent population is assumed to be normally distributed.Because the sample size is greater than 10.     Because we do not know the distribution of the parent population.Because the sample size is less than 100.