Question

# 1. There are many ways to measure the reading ability of children. Research designed to improve...

1. There are many ways to measure the reading ability of children. Research designed to improve reading performance is dependent on good measures of the outcomes. One frequently-used test is the DRP, or Degree of Reading Power. It is known that the distribution of DRP scores is normally distributed. A researcher suspects that the mean score µ of all third-graders in Henrico County Schools is different from the national mean, which is 32. To test her suspicion, she administers the DRP to a random sample of 22 Henrico County third-grade students. Their scores are recorded in the following table:

40 26 39 17 42 18 24 43 46 27 19

47 19 26 37 34 15 45 41 39 31 46

(a) Find the t critical value for a 90% confidence interval for the true mean, µ, of the population.

(b) State and verify the conditions for the confidence interval.

(c) Compute and interpret the 90% confidence interval for the mean DRP score in Henrico County Schools. (You can find your point estimate and standard error in your calculator using the 1-Var Stats function. Compare your answers with a classmate to make sure you entered things in correctly!)

(d) Use the confidence interval you constructed in Part (c) to comment on whether you agree with the researcher’s suspicion. Explain your reasoning clearly.

2. Paddy’s Pub in South Philadelphia sponsored a sample survey of 500 randomly-selected customers to get feedback on the new items they added to their menu over the summer. One question the survey asked was,“Did you try one of our new specialty drinks this summer?” Of the 500 customers who took the survey, 421 answered “Yes.”

(a) Construct and interpret a 95% confidence interval for the proportion of all Paddy’s Pub customers who tried one of the pub’s new drinks this summer. Clearly specify your z critical value.

(b) In order to promote the popularity of the new menu items, the owners at Paddy’s begin advertising that 86% of customers ordered the new specialty drinks this summer. Use the confidence interval you constructed in Part (a) to comment on whether you agree with their claim. Explain your reasoning clearly

Given Point estimate P = 421/500 = 0.842

N = 500

First we need to check the conditions of normality

That is if n*p and n*(1-p) both are greater than 5 or not

N*p = 421

And n*(1-p) = 79

As both are greater than 5

We can use standard normal z table to estimate the interval

For 95% confidence level, critical value z from z table is 1.96

Margin of error (MOE) = Z*√{p*(1-p)/n}

MOE = 0.0319709586218

Confidence interval is given by

P-MOE < P < P+MOE

0.8100290413781 < P < 0.8739709586218

B)

Yes i agree with the claim as the confidence interval contains the value 0.86 (86%).