That 46 of 100 individuals prefer Cola 2. Complete parts a through c. Do not reject the null hypothesis for the values of p0 between and inclusively. (Type integers or decimals. Do not round.)
(b) Construct a 95% confidence interval for the proportion of individuals who prefer Cola 2. We are 95% confident the proportion of individuals who prefer Cola 2 is between and (Type integers or decimals rounded to three decimal places as needed.)
(c) Suppose you changed the level of significance in conducting the hypothesis test to a=0.01. What would happen to the range of values for p0 for which the null hypothesis is not rejected? Why does this make sense? Choose the correct answer below.
A. The range of values would increase because the corresponding confidence interval would decrease in size.
B. The range of values would decrease because the corresponding confidence interval would increase in size.
C. The range of values would increase because the corresponding confidence interval would increase in size.
D. The range of values would decrease because the corresponding confidence interval would decrease in size.
The correct answer and explanation is :
Let’s go through each part step by step.
(a) Finding the Range of ( p_0 ) for Which We Do Not Reject ( H_0 )
We need to determine the values of ( p_0 ) for which we do not reject the null hypothesis. To do this, we use the confidence interval approach. Since the problem asks for values of ( p_0 ) that are not rejected, this means we look at the confidence interval, which will be calculated in part (b).
(b) Constructing a 95% Confidence Interval for ( p )
We use the formula for a confidence interval for a population proportion:
[
\hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}
]
where:
- ( \hat{p} = \frac{46}{100} = 0.46 ) (sample proportion)
- ( n = 100 ) (sample size)
- ( Z_{0.025} = 1.96 ) (critical value for a 95% confidence level)
Calculating the standard error (SE):
[
SE = \sqrt{\frac{0.46(1 – 0.46)}{100}} = \sqrt{\frac{0.46 \times 0.54}{100}} = \sqrt{0.002484} \approx 0.0498
]
Now, calculating the margin of error (ME):
[
ME = 1.96 \times 0.0498 \approx 0.0976
]
Finally, the confidence interval:
[
0.46 – 0.0976 = 0.3624, \quad 0.46 + 0.0976 = 0.5576
]
Rounded to three decimal places:
[
(0.362, 0.558)
]
So, we are 95% confident that the true proportion of individuals who prefer Cola 2 is between 0.362 and 0.558.
(c) Effect of Changing ( \alpha ) to 0.01
If we decrease the level of significance to ( \alpha = 0.01 ), this means we are using a 99% confidence interval instead of 95%. A higher confidence level means a wider confidence interval because we need to be more certain about our estimate.
Since the confidence interval increases, the range of values for ( p_0 ) for which we do not reject ( H_0 ) also increases. This corresponds to option C:
✔ C. The range of values would increase because the corresponding confidence interval would increase in size.
Explanation (300 Words)
In hypothesis testing, the level of significance (( \alpha )) determines the probability of rejecting a true null hypothesis (Type I error). When ( \alpha = 0.05 ), the critical values for a two-tailed test are based on a 95% confidence interval. If we reduce ( \alpha ) to 0.01, we are becoming more cautious and requiring stronger evidence to reject ( H_0 ), meaning our confidence level increases to 99%.
A higher confidence level leads to a wider confidence interval because we need to account for more uncertainty. This increased interval means more values of ( p_0 ) fall within the range where we do not reject the null hypothesis. Consequently, the range of ( p_0 ) values for which ( H_0 ) is not rejected also expands.
This makes sense because a smaller significance level means we are less willing to reject ( H_0 ) unless there is very strong evidence. A wider confidence interval reflects this by including a broader range of plausible values for ( p_0 ). Therefore, as the confidence level increases, the number of values of ( p_0 ) that do not lead to rejection also increases.
Thus, the correct answer is option C: The range of values would increase because the corresponding confidence interval would increase in size.