Sarah randomly selected 80 shoppers at a grocery store to taste two different types of Greek yogurt. Yogurt brand [tex]$B$[/tex] was chosen as the favorite by 53 of them.

To the nearest percent, what is the [tex]$95 \%$[/tex] confidence interval ([tex]$z^\ \textless \ em\ \textgreater \ $[/tex]-score 1.96) for the proportion of shoppers who tasted the two types of yogurt and preferred brand [tex]$B$[/tex]?

Given:
[tex]\[ E = Z^\ \textless \ /em\ \textgreater \ \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \][/tex]
[tex]\[ C = \hat{p} \pm E \][/tex]

A. between [tex]$10 \%$[/tex] and [tex]$66 \%$[/tex]

B. between [tex]$42 \%$[/tex] and [tex]$64 \%$[/tex]

C. between [tex]$52 \%$[/tex] and [tex]$81 \%$[/tex]

D. between [tex]$56 \%$[/tex] and [tex]$77 \%$[/tex]



Answer :

Let's break down the solution step by step to find the [tex]\(95 \%\)[/tex] confidence interval for the proportion of shoppers who preferred Greek yogurt brand [tex]\(B\)[/tex].

1. Determine the Sample Proportion:
- The formula for the sample proportion [tex]\(\hat{p}\)[/tex] is:
[tex]\[ \hat{p} = \frac{\text{Number of favorable outcomes}}{\text{Sample size}} \][/tex]
- From the given data, 53 out of 80 shoppers preferred brand [tex]\(B\)[/tex]:
[tex]\[ \hat{p} = \frac{53}{80} = 0.6625 \][/tex]

2. Find the Standard Error:
- The formula for the standard error (SE) is:
[tex]\[ SE = Z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \][/tex]
- Here, [tex]\(Z^*\)[/tex] is the z-score for a [tex]\(95\%\)[/tex] confidence interval, which is 1.96, [tex]\(\hat{p}\)[/tex] is 0.6625, and [tex]\(n\)[/tex] is 80:
[tex]\[ SE = 1.96 \sqrt{\frac{0.6625 \times (1 - 0.6625)}{80}} \][/tex]
- Calculate the components inside the square root:
[tex]\[ SE = 1.96 \sqrt{\frac{0.6625 \times 0.3375}{80}} \][/tex]
- Simplifying further:
[tex]\[ SE = 1.96 \sqrt{\frac{0.22359375}{80}} \][/tex]
[tex]\[ SE = 1.96 \sqrt{0.002794921875} \][/tex]
[tex]\[ SE \approx 1.96 \times 0.052857 \][/tex]
[tex]\[ SE \approx 0.1036 \][/tex]

3. Calculate the Confidence Interval:
- The confidence interval formula is:
[tex]\[ CI = \hat{p} \pm SE \][/tex]
- Using the calculated values:
[tex]\[ \hat{p} - SE = 0.6625 - 0.1036 \approx 0.5589 \][/tex]
[tex]\[ \hat{p} + SE = 0.6625 + 0.1036 \approx 0.7661 \][/tex]
- Convert these proportions to percentages:
[tex]\[ 0.5589 \times 100 \approx 55.89\% \][/tex]
[tex]\[ 0.7661 \times 100 \approx 76.61\% \][/tex]

4. State the Confidence Interval:
- Therefore, the [tex]\(95\%\)[/tex] confidence interval for the proportion of shoppers who prefer yogurt brand [tex]\(B\)[/tex] is approximately:
[tex]\[ (55.89\%, 76.61\%) \][/tex]

Among the given options, the correct range for the 95% confidence interval to the nearest percentage is:
[tex]\[ \boxed{\text{between 56 \% and 77 \%}} \][/tex]

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