A major university claimed that the mean number of credit hours that their entire population of undergraduate students took each semester was 13.1. A counselor questioned whether this was true. She took a random sample of 250 undergraduate students, and the mean of that sample of students showed that they completed 12.8 credit hours. The population standard deviation is 1.6. Conduct a full hypothesis test using the p-value approach. Let α = .05.

Determine if the mean credit hours for the sample is significantly different than that of the population.

What formula seems to match what we have been given and what we need to find?

A simple random sample of 60 items resulted in a sample mean of 70. The population standard deviation is  = 15.

a. Compute the 95% confidence interval for the population mean. Round your answers to one decimal place.
Enter your answer using parentheses and a comma, in the form (n1,n2). Do not use commas in your numerical answer (i.e. use 1200 instead of 1,200, etc.)

b. Assume that the same sample mean was obtained from a sample of 130 items. Provide a 95% confidence interval for the population mean. Round your answers to two decimal places.

c. What is the effect of a larger sample size on the interval estimate?

  Larger sample provides a - Larger or smaller - margin of error?

New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night.† Assume that room rates are normally distributed with a standard deviation of $55.

(a)

What is the probability that a hotel room costs $245 or more per night? (Round your answer to four decimal places.)

(b)

What is the probability that a hotel room costs less than $120 per night? (Round your answer to four decimal places.)

(c)

What is the probability that a hotel room costs between $210 and $300 per night? (Round your answer to four decimal places.)

You wish to test the following claim (HaHa) at a significance level of α=0.01α=0.01. For the context of this problem, μd=PostTest−PreTestμd=PostTest-PreTest where the first data set represents a pre-test and the second data set represents a post-test. (Each row represents the pre and post test scores for an individual. Be careful when you enter your data and specify what your μ1μ1 and μ2μ2 are so that the differences are computed correctly.)
Ho:μd=0
Ha:μd≠0
You believe the population of difference scores is normally distributed, but you do not know the standard deviation. You obtain the following sample of data:

What is the test statistic for this sample?
test statistic =

What is the p-value for this sample?
p-value =

LOADING...

Click the icon to view the table of critical​ t-values.

​(a) Determine a point estimate for the population mean.

A point estimate for the population mean is BLANK

​(Round to two decimal places as​ needed.)

​(b) Construct and interpret a 95​% confidence interval for the mean pH of rainwater. Select the correct choice below and fill in the answer boxes to complete your choice.

​(Use ascending order. Round to two decimal places as​ needed.)

A. There is 95​% confidence that the population mean pH of rain water is between BLANK AND BLANK

B. If repeated samples are​ taken, 95​% of them will have a sample pH of rain water between BLANK and BLANK.

C. There is a 95​% probability that the true mean pH of rain water is between BLANK AND BLANK.

​(c) Construct and interpret a 99​% confidence interval for the mean pH of rainwater.

Select the correct choice below and fill in the answer boxes to complete your choice.

​(Use ascending order. Round to two decimal places as​ needed.)

A.There is 99​% confidence that the population mean pH of rain water is between BLANK AND BLANK.

B. If repeated samples are​ taken, 99​% of them will have a sample pH of rain water between BLANK AND BLANK.

C. There is a 99​% probability that the true mean pH of rain water is between BLANK and BLANK.

​(d) What happens to the interval as the level of confidence is​ changed? Explain why this is a logical result.

As the level of confidence​ increases, the width of the interval

▼

decreases.

increases.

This makes sense since the

▼

sample size

margin of error

point estimate

▼

decreases as well.

increases as well.

A medical journal reported the results of a study in which three groups of 50 women were randomly selected and monitored for urinary tract infections over 6 months. One group drank cranberry juice​ daily, one group drank a lactobacillus​ drink, and the third group drank neither of those​ beverages, serving as a control group. In the control​ group, 18 women developed at least one infection compared with 19 of those who consumed the lactobacillus drink and only 7 of those who drank cranberry juice. Does the study provide supporting evidence for the value of cranberry juice in warding off urinary tract infections in​ women? Complete parts a through f below.

f) If you concluded that the groups are not the​ same, analyze the differences using the standardized residuals of your calculations. Select the correct choice​ below, and if​ necessary, fill in the answer box to complete your choice.

A.

B. The conclusion does not indicate that the groups are different.

C. Since the assumptions are not​ satisfied, a hypothesis test is not appropriate.

3. Assume that oak trees have an average height of 90 feet with a standard

deviation of 14 feet. Their heights are normally distributed (i.e., μ = 90 and σ = 14).

A. Using a z table or online calculator, determine the percent of oak trees that are at least 106.50 feet tall. (Hint: You will need to start by converting 106.50 to a z score.)

B. Using a z table or online calculator, determine the percent of oak trees that are 83.95 feet or less.

C. Using your answers to A and B, what percent of oak trees’ heights are between 83.95 feet and 106.50 feet?

what is the difference between one sample t test ,two sample test and paired test? can you make a real life example to illustrate?

A sample of 35 cars of a certain kind had an average mileage of 36.2 mpg. Assuming that mileage is approximately normally distributed with standard deviation 4 mpg, test the hypothesis that the average mileage for all cars of this type is no less than 34.2 mpg at the 0.01 significance level. Give the value of p you find to two decimal places, and choose the correct conclusion:

p=

Problems 1-4 assume a normally distributed population with a mean = 48 and standard deviation = 5. Be sure to sketch the curve, include formulas & work, round appropriately, and circle your final answer.

1. What percent of the scores fall:

at or below 54?

at or above 40?

2. (6 points) What proportion of scores lie between:

31 and 48?

31 and 54?

3. (4 points) Suppose we took a sample of 5000 scores from this distribution.

Of those 5000 scores, how many would you expect to lie between31 and 48?

Of those 5000 scores how many would you expect to lie between 31 and 54?

4. If you select one person at random from this population and find his/her score on this variable to be 63 would you be surprised (yes/no)? Why/why not?

5. (4 points) Later in the semester we will focus on z-scores that capture the middle 95% of the distribution.
   1. Find the two z-scores that capture the middle 95% of the distribution. Be sure to show your work. (Hint: remember that the normal curve is symmetrical.)

What two z-scores would we use if we want to capture the middle 99% of the distribution?

(I) You may consider using function prop.test to perform the above hypothesis test.

(II) Present complete procedures of hypothesis testing for the above problem such as null hypothesis, alternative hypothesis, significance level, test statistics value, p-value etc..in your findings.

(III) State your conclusion clearly.

A cell-phone store sold 150 smartphones of Brand A and 14 of them returned as defective items. Besides that, the cell-phone store sold also 125 smartphones of Brand B and 15 phones of them retuned as defective items. Is there any statistical evidence that Brand A has a smaller chance of being returned than Brand B at:

(i) 1% significance level?

(ii) 5% significance level?

(iii) 10% significance level?

Justify your findings.

The overhead reach distances of adult females are normally distributed with a mean of

202.5 cm202.5 cm

and a standard deviation of

8.6 cm8.6 cm.

a. Find the probability that an individual distance is greater than

215.90215.90

cm.

b. Find the probability that the mean for

2020

randomly selected distances is greater than 200.70 cm.200.70 cm.

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The types of raw materials used to construct stone tools found at an archaeological site are shown below. A random sample of 1486 stone tools were obtained from a current excavation site.

Use a 1% level of significance to test the claim that the regional distribution of raw materials fits the distribution at the current excavation site.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: The distributions are different.
H₁: The distributions are different.

H₀: The distributions are different.
H₁: The distributions are the same.     

H₀: The distributions are the same.
H₁: The distributions are different.

H₀: The distributions are the same.
H₁: The distributions are the same.

(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?

Yes or No     

What sampling distribution will you use?

chi-square

normal     

uniform

Student's t

binomial

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.)


(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 0.01 level of significance, there is sufficient evidence to conclude that the regional distribution of raw materials and the current excavation site distribution are not independent.

At the 0.01 level of significance, there is insufficient evidence to conclude that the regional distribution of raw materials and the current excavation site distribution are not independent.     

Gas Mileage. Based on tests of the Chevrolet Cobalt, engineers have found that the miles per gallon in highway driving are normally distributed, with a mean of 32 MPG and a standard deviation of 3.5 MPG.

a) What is the probability that a randomly selected Cobalt gets more than 34 MPG?

b) Suppose that 10 Cobalts are randomly selected and the MPG for each car are recorded. What is the probability that the mean MPG exceeds 34 MPG?

c) Suppose 20 Cobalts are randomly selected and the MPG for each car are recorded. What is the probability that the mean MPG exceeds 34 MPG?

QUESTION 1

1. A random group of oranges were selected from an orchard to analyze their ripeness. The data is shown

   below:

   Ready to pick

   Ripe

   Ready in three days

   Ready in one week

   Ready in two weeks

   Number of oranges

   11

   20

   13

   17

   Based on the time of year, the orchard owner believes that 30% of the oranges are ready for picking now,

   30% will be ready in three days, 30% will be ready in one week, and 10% will be ready in two weeks.
   Is there evidence to reject this hypothesis at a  .05 significance level?  

   		There is evidence to reject the claim that the oranges are distributed as claimed because the test value 24.082 > 9.488
   		There is not evidence to reject the claim that the oranges are distributed as claimed because the test value 5.991 < 24.082
   		There is evidence to reject the claim that the oranges are distributed as claimed because the test value 24.082 > 7.815
   		There is not evidence to reject the claim that the oranges are distributed as claimed because the test value 9.488 < 24.082