Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ² = 5.1. Suppose a recent study of age at first marriage for a random sample of 51 women in rural Quebec gave a sample variance s² = 3.0. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence interval for the population variance. (a) What is the level of significance?


State the null and alternate hypotheses.

H_o: σ² = 5.1; H₁: σ² ≠ 5.1

H_o: σ² = 5.1; H₁: σ² > 5.1    

H_o: σ² < 5.1; H₁: σ² = 5.1

H_o: σ² = 5.1; H₁: σ² < 5.1

(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)


What are the degrees of freedom?


What assumptions are you making about the original distribution?

We assume a normal population distribution.

We assume a uniform population distribution.    

We assume a binomial population distribution.

We assume a exponential population distribution.

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

P-value > 0.100

0.050 < P-value < 0.100    

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

P-value < 0.005

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

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 the variance of age at first marriage is less than 5.1.

At the 5% level of significance, there is sufficient evidence to conclude that the that the variance of age at first marriage is less than 5.1.    

(f) Find the requested confidence interval for the population variance. (Round your answers to two decimal places.)

Interpret the results in the context of the application.

We are 90% confident that σ² lies outside this interval.

We are 90% confident that σ² lies below this interval.    

We are 90% confident that σ² lies within this interval.

We are 90% confident that σ² lies above this interval.

D) Use the normal model to determine the proportion of babies in each class

How do I manually determine the normal mode? Please provide step by step manually (excel is what I am using, however I need to show steps.

Thank you.

6.20. A convenience store owner wants to know how long his customers spend browsing the store before making a purchase. It is found that time spent is normally distributed with an average of m = 5 minutes and a standard deviation of s=2 : 2 minutes. Using a random sample of 14 customers, what is the probability that a customer, on average, will spend less than 4 minutes browsing the store?

1a) Explain why we reject the null hypothesis when the p-value is less than the level of significance?

b) Explain to someone unfamiliar with statistics how to tell whether a statistical test is left, right, or two tailed. Explain what to look for in the wording of a hypothesis test and with the alternate hypothesis.

c) Why can we never truly accept the null hypothesis?

Question 4:

The times that a cashier spends processing each person’s transaction are independent and identically distributed random variables with a mean of µ and a variance of σ² . Thus, if X_i is the processing time for each transaction, E(X _i) = µ and Var(X_i) = σ² .

Let Y be the total processing time for 100 orders: Y = X₁ + X₂ + · · · + X₁₀₀

(a) What is the approximate probability distribution of Y , the total processing time of 100 orders? Hint: Y = 100X, where X = 1 100 P100 i=1 Xi is the sample mean.

(b) Suppose for Z ∼ N(0, 1), a standard normal random variable:

P(a < Z < b) = 100(1 − α)%

Using your distribution from part (a), show that an approximate 100(1 − α)% confidence interval for the unknown population mean µ is:

(Y − 10bσ)/100 < µ < (Y − 10aσ)/100

(c) Now suppose that the population mean processing time is known to be µ = 1.5 minutes, and the population standard deviation processing time is known to be σ = 1 minute. What is the probability that it takes less than 120 minutes to process the 100 orders? If you use R, please provide the commands used to determine the probability. Could you show all steps in the hand written working for this question please.

6. Digital streaming has shifted some of the focus from traditional TV commercials to online advertisements.
The Harris poll reported in 2012 that 53% of 2,343 American adults surveyed said they watched digitally
streamed content. Calculate and interpret a 99% score CI for the proportion of all adult Americans
who watched digitally streamed content.

A husband and wife, Ed and Rina, share a digital music player that has a feature that randomly selects which song to play. A total of 3476 songs have been loaded into the player, some by Ed and the rest by Rina. They are interested in determining whether they have each loaded different proportions of songs into the player. Suppose that when the player was in the random-selection mode, 38 of the first 58 songs selected were songs loaded by Rina. Let p denote the proportion of songs that were loaded by Rina.

(a) State the null and alternative hypotheses to be tested. How strong is the evidence that Ed and Rina have each loaded a different proportion of songs into the player? Make sure to check the conditions for the use of this test. (Round your test statistic to two decimal places and your P-value to three decimal places. Assume a 95% confidence level.)

z = 2.36

P-value = 0.018

Conclusion: There is strong evidence that the proportion of songs downloaded by Ed and Rina differs from 0.5.

(b) Are the conditions for the use of the large sample confidence interval met? Yes, the conditions are met.

If so, estimate with 95% confidence the proportion of songs that were loaded by Rina. Round your answers to 3 decimal places. _____ to ________

A survey of 1935 people who took trips revealed that 181 of them included a visit to a theme park. Based on those survey results, a management consultant claims that less than 10 % of trips include a theme park visit. Test this claim using the ?=0.01 significance level.

(a) The test statistic is

(b) The P-value is

(c) The conclusion is  

A. There is not sufficient evidence to support the claim that less than 10 % of trips include a theme park visit.
B. There is sufficient evidence to support the claim that less than 10 % of trips include a theme park visit.

A random sample of ? measurements was selected from a population with standard deviation σ=11.7 and unknown mean μ. Calculate a 90 % confidence interval for μ for each of the following situations:

(a) ?=40, ?=72.1
≤. μ ≤

(b)  ?=60, ?=72.1
μ ≤

(c)  ?=85, ?=72.1
≤. μ. ≤

Construct an 80​% confidence interval to estimate the population mean when x overbar=131 and s​ = 28 for the sample sizes below. ​

a) n=20

​b) n=50 ​

c) n=80

Tom Marley and Jennifer Griggs have recently started a marketing research firm in Jacksonville, Florida. They have contacted the Florida Democratic Party with a proposal to do all political polling for the party. Since they have just started their company, the state party chairman is reluctant to sign a contract without some test of their accuracy. He has asked them to do a trial poll in a central Florida county known to have 60% registered Democratic Party voters. The poll itself had many questions. However, for the test of accuracy, only the proportion of registered Democrats was considered. Tom and Jennifer report back that from a random sample of 1000 respondents, 520 were registered Democrats.

1. Determine the probability that such a random sample would result in 520 or fewer Democrats in the sample.

2. Based on your calculations in part a, would you recommend that the Florida Democratic Party (or anyone else for that matter) contract with the Marley/Griggs marketing research firm? Explain your answer.

Hemoglobin (g/100mL) was measured twice in 20 pregnant women. A first measurement was taken at 1-3 weeks prepartum whereas the second measurement was taken at 2-6 days postpartum (the data is provided below).

Part A. What is the absolute value of the t-test statistic for testing whether there is any change in mean hemoglobin levels between prepartum and postpartum women?

Part B. What is the degrees of freedom for this particular test in Part A?

a. 20

b. 38

c. 19

d. 25.6

Part C. What would you conclude from the corresponding t-test?

a. Mean hemoglobin decreases from prepartum to postpartum stages in women (P=0.02)

b. Mean hemoglobin decreases from prepartum to postpartum stages in women (P=0.04)

c. There is no evidence of a mean difference between prepartum and postpartum stages (P>0.05)

Part D. Suppose a hospital will not pursue any intervention measures if they believe that the true change in hemoglobin levels between prepartum and postpartum stages falls within +/- 2g/100mL; in other words, they feel the two stages are effectively equivalent for mean hemoglobin levels if the true mean difference falls within +/- 2g/100mL. Does this dataset provide proof of equivalence using a = 0.05?  

a. Yes based on a confidence interval of [0.18,186]

b. Yes based on a confidence interval of [0.20,1.84]

c. No based on a confidence interval of [0.32,1.72]

d. No based on a confidence interval of [0.03,2.01]

e. Yes based on a confidence interval of [0.32,1.72]

Data:

Please state which of the following apply to questions

6. Pr(r = 6 | n = 22, p = 48%)

Binomial

Poisson

Hypergeometric

None of the above

7. Pr(r = 6 | n = 13, p = 52%)

Binomial

Poisson

Hypergeometric

None of the above

8. N = 47

Binomial

Poisson

Hypergeometric

None of the above

9. The probability of flying from New York to Paris in under 7 hours, 10 minutes

Binomial

Poisson

Hypergeometric

None of the above

10. n = 100, p = 4.2%

Binomial

Poisson

Hypergeometric

None of the above

7. A data set includes 108 body temperatures of healthy adult humans having a mean of 98.3 F° and a standard deviation of 0.69 F°. Construct a 99​% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6 F° as the mean body​ temperature?

What is the confidence interval estimate of the population mean µ​?

____F°<µ<____ F°

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

What does this suggest about the use of 98.6 F° as the mean body​ temperature?

A.This suggests that the mean body temperature could be lower than 98.6 F°.

B.This suggests that the mean body temperature could be higher than 98.6 F°

C.This suggests that the mean body temperature could very possibly be 98.6 F°.

8. An IQ test is designed so that the mean is 100 and the standard deviation is 14 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 95​% confidence that the sample mean is within 7 IQ points of the true mean. Assume that σ= 14 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.

The required sample size is

nothing. ​(Round up to the nearest​ integer.)

Would it be reasonable to sample this number of​ students?

Yes. This number of IQ test scores is a fairly small number.

Yes. This number of IQ test scores is a fairly large number.

No. This number of IQ test scores is a fairly large number.

No. This number of IQ test scores is a fairly small number

Use the ”test.vs.grade” data and test the null hypothesis that the mean test score for the population is 70 against the alternative that it is greater than 70. Find a p-value and state your conclusion if α = 0.05. Repeat for the null hypothesis µ = 75.

https://www.math.uh.edu/~charles/data/test.vs.grade.csv