Question

A survey of Internet users reported that 17% downloaded music onto their computers. The filing of lawsuits by the recording industry may be a reason why this percent has decreased from the estimate of 27% from a survey taken two years before. Suppose we are not exactly sure about the sizes of the samples. Perform the calculations for the significance tests and 95% confidence intervals under each of the following assumptions. (Use previous − recent. Round your test statistics to two decimal places and your confidence intervals to four decimal places.)

(i) Both sample sizes are 1000.

z =

95% C.I. ( , )

(ii) Both sample sizes are 1600.

z =

95% C.I. ( , )

(iii) The sample size for the survey reporting 27% is 1000 and the sample size for the survey reporting 17% is 1600.

z =

95% C.I. ( , )

Summarize the effects of the sample sizes on the results.

We see in (i) and (ii) that smaller samples result in larger z (weaker evidence) and smaller intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated.

We see in (i) and (ii) that smaller samples result in smaller z (stronger evidence) and wider intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated.

We see in (i) and (ii) that smaller samples result in smaller z (weaker evidence) and wider intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated.

We see in (i) and (ii) that smaller samples result in larger z (stronger evidence) and narrower intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated.

We see in (i) and (ii) that smaller samples result in smaller z (weaker evidence) and narrower intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated.

Answer 1

Solution:-

(i) 95% confidence intervals for the difference in proportion is C.I = (- 0.136, 0.064).

Both sample sizes are 1000.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P₁ = P₂
Alternative hypothesis: P₁ P₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)
p = 0.22
SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.01853
z = (p₁ - p₂) / SE

z = - 5.39

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 5.39 or greater than 5.39.

Thus, the P-value = less than 0.0001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is a significance difference in the proportions.

C.I = (0.17 - 0.27) + 1.96*0.01853

C.I = - 0.10 + 0.03632

C.I = (- 0.136, 0.064)

(ii) 95% confidence intervals for the difference in proportion is C.I = (- 0.129, 0.0714).

Both sample sizes are 1600.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P₁ = P₂
Alternative hypothesis: P₁ P₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)
p = 0.22
SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.01465
z = (p₁ - p₂) / SE

z = - 6.83

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 6.83 or greater than 6.83.

Thus, the P-value = less than 0.0001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is a significance difference between proportions.

C.I = (0.17 - 0.27) + 1.96*0.01465

C.I = - 0.10 + 0.02862

C.I = (- 0.129, 0.0714)

(iii) 95% confidence intervals for the difference in proportion is C.I = (- 0.1318, 0.0682).

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P₁ = P₂
Alternative hypothesis: P₁ P₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)
p = 0.22
SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.0167
z = (p₁ - p₂) / SE

z = - 5.99

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 5.99 or greater than 5.99.

Thus, the P-value = less than 0.0001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is a significance difference in proportions.

C.I = (0.17 - 0.27) + 1.96*0.01622

C.I = - 0.10 + 0.0318

C.I = (- 0.1318, 0.0682)

d) We see in (i) and (ii) that smaller samples result in larger z (stronger evidence) and narrower intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated.