Twenty-nine college students, identified as having a positive attitude about Mitt Romney as compared to Barack Obama in the 2012 presidential election, were asked to rate how trustworthy the face of Mitt Romney appeared, as represented in their mental image of Mitt Romney’s face. Ratings were on a scale of 0 to 7, with 0 being “not at all trustworthy” and 7 being “extremely trustworthy.” Here are the 29 ratings: 2.63.23.73.33.43.63.73.83.94.14.24.95.74.23.93.24.55.05.04.64.63.93.95.32.82.63.03.33.7 a 95% confidence interval for the mean rating.Is there significant evidence at the 5% level that the mean rating is greater than 3.5 (a neutral rating)?

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Asked Mar 19, 2019
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Twenty-nine college students, identified as having a positive attitude about Mitt Romney as compared to Barack Obama in the 2012 presidential election, were asked to rate how trustworthy the face of Mitt Romney appeared, as represented in their mental image of Mitt Romney’s face. Ratings were on a scale of 0 to 7, with 0 being “not at all trustworthy” and 7 being “extremely trustworthy.” Here are the 29 ratings: 

2.6 3.2 3.7 3.3 3.4 3.6 3.7 3.8 3.9 4.1
4.2 4.9 5.7 4.2 3.9 3.2 4.5 5.0 5.0 4.6
4.6 3.9 3.9 5.3 2.8 2.6 3.0 3.3 3.7  
  1. a 95% confidence interval for the mean rating.
  2. Is there significant evidence at the 5% level that the mean rating is greater than 3.5 (a neutral rating)?
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Expert Answer

Step 1

a.

Here, the sample size is 29 that is reasonably small. The population standard deviation is unknown here. Therefore, z-test cannot be used here. In this situation student’s t-distribution can be used.

Let µ be the true mean rating.

Let x bar be the sample mean rating. s be the sample standard deviation. n is the sample size.

The 100(1-α)% confidence interval of µ is,

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Step 2

Here the confidence interval is 95%.

Therefore, α = 0.05.

The degrees of freedom, n - 1 = 29 - 1 =28,

From the EXCEL, using the formula, =T.INV.2T(0.05,28), the critical value is 2.0484.

From the EXCEL, the mean of the data is 3.9172 and the standard deviation is 0.796.

Therefore the 95% confidence interval of mean rating is,

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Step 3

b.

State the appropriate null and alternative hypotheses:

Null hypothesis:

H0 : μ = 3.5

That is, the mean rating is 3.5.

Alternative hypothesis:

H1 : μ > 3.5

That is, the mean rating is more than ...

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