7.2 hw hypothesis tests, proportions 10 questions 2

Statsmath

1. You are conducting a study to see if the probability of a true negative on a test for a certain cancer is significantly more than 0.25.
With H1 : p >> 0.25 you obtain a test statistic of z=1.397z=1.397.
Use a normal distribution calculator and the test statistic to find the P-value accurate to 4 decimal places. It may be left-tailed, right-tailed, or 2-tailed.

P-value =

2. You are conducting a study to see if the probability of catching the flu this year is significantly more than 0.27.
With H1 : p >> 0.27 you obtain a test statistic of z=1.722z=1.722.
Use a normal distribution calculator and the test statistic to find the P-value accurate to 4 decimal places. It may be left-tailed, right-tailed, or 2-tailed.

P-value = 

3. You are conducting a study to see if the probability of a true negative on a test for a certain cancer is significantly more than 0.81. You use a significance level of α=0.001α=0.001.

H0:p=0.81H0:p=0.81
 H1:p>0.81H1:p>0.81

You obtain a sample of size n=218n=218 in which there are 184 successes.

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

The p-value is…

  • a) less than (or equal to) αα
  • b) greater than αα

This test statistic leads to a decision to…

  • a) reject the null
  • b) accept the null
  • c) fail to reject the null

As such, the final conclusion is that…

  • a) There is sufficient evidence to warrant rejection of the claim that the probability of a true negative on a test for a certain cancer is more than 0.81.
  • b)There is not sufficient evidence to warrant rejection of the claim that the probability of a true negative on a test for a certain cancer is more than 0.81.
  • c)The sample data support the claim that the probability of a true negative on a test for a certain cancer is more than 0.81.
  • d)There is not sufficient sample evidence to support the claim that the probability of a true negative on a test for a certain cancer is more than 0.81.

4. You are conducting a study to see if the proportion of men over 50 who regularly have their prostate examined is significantly different from 0.23. You use a significance level of α=0.02α=0.02.

H0:p=0.23H0:p=0.23
 H1:p≠0.23H1:p≠0.23

You obtain a sample of size n=167n=167 in which there are 32 successes.

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

The p-value is…

  • A) less than (or equal to) αα
  • B) greater than αα

This test statistic leads to a decision to…

  • A)reject the null
  • B)accept the null
  • C)fail to reject the null

As such, the final conclusion is that…

  • A) There is sufficient evidence to warrant rejection of the claim that the proportion of men over 50 who regularly have their prostate examined is different from 0.23.
  • B) There is not sufficient evidence to warrant rejection of the claim that the proportion of men over 50 who regularly have their prostate examined is different from 0.23.
  • C) The sample data support the claim that the proportion of men over 50 who regularly have their prostate examined is different from 0.23.
  • D) There is not sufficient sample evidence to support the claim that the proportion of men over 50 who regularly have their prostate examined is different from 0.23.

5. Many investors and financial analysts believe the Dow Jones Industrial Average (DJIA) gives a good barometer of the overall stock market. On January 31, 2006, 9 of the 30 stocks making up the DJIA increased in price (The Wall Street Journal, February 1, 2006). On the basis of this fact, a financial analyst claims we can assume that 30% of the stocks traded on the New York Stock Exchange (NYSE) went up the same day.

A sample of 75 stocks traded on the NYSE that day showed that 26 went up.

You are conducting a study to see if the proportion of stocks that went up is significantly more than 0.3. You use a significance level of α=0.001α=0.001.

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

The p-value is…

  • A) less than (or equal to) αα
  • B) greater than αα

This test statistic leads to a decision to…

  • A) reject the null
  • B) accept the null
  • C) fail to reject the null

As such, the final conclusion is that…

  • A)There is sufficient evidence to warrant rejection of the claim that the proportion of stocks that went up is more than 0.3.
  • B)There is not sufficient evidence to warrant rejection of the claim that the proportion of stocks that went up is more than 0.3.
  • C)The sample data support the claim that the proportion of stocks that went up is more than 0.3.
  • D)There is not sufficient sample evidence to support the claim that the proportion of stocks that went up is more than 0.3.

6. A well-known brokerage firm executive claimed that 60% of investors are currently confident of meeting their investment goals. An XYZ Investor Optimism Survey, conducted over a two week period, found that in a sample of 400 people, 54% of them said they are confident of meeting their goals.

Test the claim that the proportion of people who are confident is smaller than 60% at the 0.005 significance level.

The null and alternative hypothesis would be:

A) H0:p≥0.5H0:p≥0.5
H1:p<0.5H1:p<0.5

B) H0:μ≤0.5H0:μ≤0.5
H1:μ>0.5

C) H0:μ=0.5H0:μ=0.5
H1:μ≠0.5

D) H0:μ≥0.5H0:μ≥0.5
H1:μ<0.5H1:μ<0.5

E) H0:p=0.5H0:p=0.5
H1:p≠0.5

F) H0:p≤0.5H0:p≤0.5
H1:p>0.5H1:p>0.5

The test is:

A) two-tailed
B) left-tailed
C) right-tailed

The test statistic is:  (to 3 decimals)

The p-value is:  (to 4 decimals)

Based on this we:

  • A) Fail to reject the null hypothesis
  • B) Reject the null hypothesis

7. Test the claim that the proportion of people who own cats is significantly different than 50% at the 0.1 significance level.

The null and alternative hypothesis would be:

A) H0:p≥0.5H0:p≥0.5
H1:p<0.5H1:p<0.5

B) H0:μ≤0.5H0:μ≤0.5
H1:μ>0.5H1:μ>0.5

C) H0:μ=0.5H0:μ=0.5
H1:μ≠0.5H1:μ≠0.5

D) H0:μ≥0.5H0:μ≥0.5
H1:μ<0.5H1:μ<0.5

E) H0:p=0.5H0:p=0.5
H1:p≠0.5H1:p≠0.5

F) H0:p≤0.5H0:p≤0.5
H1:p>0.5H1:p>0.5

The test is:

A) left-tailed
B) two-tailed
C)right-tailed

Based on a sample of 300 people, 44% owned cats

The test statistic is:             (to 2 decimals)

The p-value is:          (to 2 decimals)

Based on this we:

  • A) Reject the null hypothesis
  • B) Fail to reject the null hypothesis

8. Given ˆpp^ = 0.2571 and N = 35 for the high-income group,

Test the claim that the proportion of children in the high-income group that drew the nickel too large is smaller than 50%. Test at the 0.01 significance level.

a) Identify the correct alternative hypothesis:

  • A)    μ=.50μ=.50
  • B)    p>.50p>.50
  • C)    μ<.50μ<.50
  • D)    p=.50p=.50
  • E)    μ>.50μ>.50
  • F)    p<.50p<.50

Give all answers correct to 3 decimal places.

b) The test statistic value is:   

c) Using the P-value method, the P-value is:

d) Based on this, we

  • A) Reject H0H0
  • B) Fail to reject H0H0

e) Which means

  • A) There is not sufficient evidence to warrant rejection of the claim
  • B) There is sufficient evidence to warrant rejection of the claim
  • C) There is not sufficient evidence to support the claim
  • D) The sample data support the claim

9. A well-known brokerage firm executive claimed that 70% of investors are currently confident of meeting their investment goals. An XYZ Investor Optimism Survey, conducted over a two-week period, found that in a sample of 100 people, 64% of them said they are confident of meeting their goals.

Test the claim that the proportion of people who are confident is smaller than 70% at the 0.10 significance level.

The null and alternative hypothesis would be:

A)   H0:p≤0.7H0:p≤0.7
H1:p>0.7H1:p>0.7

B)   H0:μ=0.7H0:μ=0.7
H1:μ≠0.7H1:μ≠0.7

C)    H0:μ≤0.7H0:μ≤0.7
H1:μ>0.7H1:μ>0.7

D)   H0:μ≥0.7H0:μ≥0.7
H1:μ<0.7H1:μ<0.7

E)    H0:p≥0.7H0:p≥0.7
H1:p<0.7H1:p<0.7

F).   H0:p=0.7H0:p=0.7
H1:p≠0.7H1:p≠0.7

The test is:

A)  right-tailed
B)  left-tailed
C)  two-tailed

The test statistic is:                  (to 3 decimals)

The p-value is:                          (to 4 decimals)

Based on this we:

  • A) Reject the null hypothesis
  • B) Fail to reject the null hypothesis

10. The US Department of Energy reported that 50% of homes were heated by natural gas. A random sample of 349 homes in Oregon found that 129 were heated by natural gas. Test the claim that proportion of homes in Oregon that were heated by natural gas is different than what was reported. Use a 1% significance level. Give answer to at least 4 decimal places.

What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.)

H0: Select an answer s² x̄ σ² p p̂ μ σ s  ? = ≤ ≥ ≠ > <  

H1: Select an answer σ² s x̄ σ μ p̂ s² p  ? ≥ ≤ = ≠ < >   

Based on the hypotheses, compute the following:

Test Statistic =

p-value = 

Based on the above we choose to Select an answer Accept the alternative hypothesis Reject the null hypothesis Accept the null hypothesis Fail to reject the null hypothesis 

The correct summary would be: Select an answer There is not enough evidence to reject the claim There is enough evidence to support the claim There is not enough evidence to support the claim There is enough evidence to reject the claim that the proportion of homes in Oregon that were heated by natural gas is different than what the DOE reported value of 50%.

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