Question

The population of all verbal GRE scores are known to have a standard deviation of 8.5. The UW Psychology department hopes to receive applicants with a verbal GRE scores over 210. This year, the mean verbal GRE scores for the 42 applicants was 212.79. Using a value of α = 0.05 is this new mean significantly greater than the desired mean of 210?

a. Is this a one tailed or two tailed test? Why?

b. Conduct the appropriate hypothesis test to determine whether the new mean is significantly greater, as hypothesised.

c. If the researcher had used a sample of n = 20 students and obtained the same sample mean, would the results be sufficient to reject H0?

d. What does this say about the relevance of a sample size in hypothesis testing. Refer to the two samples used above.

e. Suppose the average caloric count of Asian food is normally distributed with a standard deviation of 1. Just for fun, you sample 58 Asian foods from your favourite restaurant and obtain a mean caloric count of 58.03 and a standard deviation of 0.9394. Using an alpha value of α = 0.01, is this observed mean significantl different from the expected caloric count of 58?

Answer 1

Data given to us is as follows:

Standard Deviation, S = 8.5

Sample mean, m = 212.79

Sample size, n = 42

Significance level, a = 0.05

(a)

This is a one tailed hypothesis test, specifically, this is a right tailed hypothesis test, because here the population mean is checked to be greater than the value 210.

(b)

The hypotheses are:

H0: = 210

Ha: > 210

Standard error is calculated as:

SE = S/(n^0.5) = 8.5/(42^0.5) = 1.31

Calculate the z-statistic:

z = (m-210)/SE = (212.79-210)/1.31 = 2.129

The corresponding p-value for this z-value is:

p = 0.016

Since p < a. we have to reject the null hypothesis.

So, the new mean is significantly greater, as hypothesised.

(c)

In this case:

Standard error is calculated as:

SE = S/(n^0.5) = 8.5/(20^0.5) = 1.9

Calculate the z-statistic:

z = (m-210)/SE = (212.79-210)/1.9 = 1.468

The corresponding p-value for this z-value is:

p = 0.071

Since p > a. we can't reject the null hypothesis.

So, the new mean is not significantly greater, as hypothesised.

(d)

As sample size increases, the standard error of the mean also increases, which decreases the chances of the data obtained as being significant.

(e)

Data given to us is as follows:

Standard Deviation, S = 1

Sample mean, m = 58.03

Sample size, n = 58

Significance level, a = 0.01

The hypotheses are:

H0: = 58

Ha: 58

Standard error is calculated as:

SE = S/(n^0.5) = 1/(58^0.5) = 0.131

Calculate the z-statistic:

z = (m-58)/SE = (58.03-58)/0.131 = 0.229

The corresponding p-value for this two tailed test using this z-value is:

p = 0.81

Since p > a. we can't reject the null hypothesis.

So, the observed mean is not significantly different from the expected mean of 58.