Question

Metacritic is a website that aggregates reviews of music, games, and movies. For each product, a numerical score is obtained from each review and the website posts the average core as well as individual reviews. The website is somewhat similar to Rotten Tomatoes, but Metacritic uses a different method of scoring that converts each review into score in 100-point scale. In addition to using the reviewers quantitative ratings (stars, 10-point scale), Metacritic manually assesses the tone of the review before scoring. Historical data shows that these converted scores are normally distributed. One of the movies that the Metacritic rated was Zootopia. Based on the data from Metacritic on November 20, 2017, there are n=43 reviews, the sample average score is 77.86, and the sample standard deviation is 11.30.

A 95% confidence interval for the true average score (µ) of Zootopia is:

a) [75.21, 81.50]
b) [76.38, 80.34]
c) [77.15, 82.84]
d) [78.96, 81.76]
e) None of the above

Answer 1

`77.86-2.02(11.30)/(√(43))=74.38`

`77.86+2.02(11.30)/(√(43))=81.34`

And for this cae none of the options satisfy the result so then the best option would be:

e) None of the above

Explanation:

Information given by the problem

`\bar X= 77.86` represent the sample mean for the score

`\mu` population mean

s=11.30 represent the sample standard deviation

n=43 represent the sample size

Calculating the confidence interval

The confidence interval for the true mean of interest is given by:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

For this case the degrees of freedom are:

`df=n-1=43-1=42`

The Confidence is 0.95 or 95%, the significance then is
`\alpha=1-0.95=0.05` and
`\alpha/2 =0.025`, the critical value for this case would be
`t_(\alpha/2)=2.02`

And replacing in equation (1) we got:

`77.86-2.02(11.30)/(√(43))=74.38`

`77.86+2.02(11.30)/(√(43))=81.34`

And for this cae none of the options satisfy the result so then the best option would be:

e) None of the above

Answer 2

e) None of the above

Explanation:

We are in posession of the sample's standard deviation, so we use the student t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 43 - 1 = 42

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 42 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.95)/(2) = 0.975`. So we have T = 2.0181

The margin of error is:

M = T*s = 2.0181*11.3 = 22.80

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 77.86 - 22.8 = 55.06

The upper end of the interval is the sample mean added to M. So it is 77.86 + 22.8 = 100.66.

So the correct answer is:

e) None of the above