MATH 2565 - Winter 2019 Assignment 4 Question 1: Sample size, n

MATH 2565 - Winter 2019
Assignment 4
Question 1:
Sample size, n = 560 ; sample proportion, ;
a. The null and alternative hypothesis are defined as:
The critical value at is: (left-tailed test)
The test statistic is calculated as:
The test statistic is less than the critical value (-2.732 ≤ -2.326), we reject the null
hypothesis. Therefore, there is sufficient statistical evidence to suggest that the true
proportion of residents who favor additional power to tap phones is less than 0.75.
b. The p-value associated with this hypothesis test is: p = 0.0031; hence p-value <
0.01
Question 2:
Sample size, n = 15; sample mean, = 3.63; sample standard deviation, s = 1.67; α =
0.05
The null and alternative hypothesis are defined as:
Since the sample size is small (n<30), the appropriate test statistic is:
This is a right tailed test, using a t test statistics and 5% significance level. The critical
value at and degrees of freedom, df= 15-1=14 is:
The decision rule is to Reject if t ≥ 2.145

The test statistic is calculated as:
The test statistic, T= 1.461 is less than the critical value, t =2.145, we fail to reject the
null hypothesis. Therefore, there is no statistically sufficient evidence at to
suggest that the true population mean FEF is greater than 3.0.
Question 3:
Historical mean thickness = 3.32 meters; sample size, n = 28; sample mean, = 4.036
meters; σ = 2.8 meters; assuming it’s normally distributed.
a. The null and alternative hypothesis are defined as:
Since the sample is normally distributed, the appropriate test statistic is: .
This is a two tailed test, using a z test statistics and 1% significance level. The critical
value of t is: . The decision rule is to Reject if z ≤ -2.576 or z ≥
2.576.
The test statistic is computed as:
The conclusion is not to reject the null hypothesis because 1.353 ≤ 2.771. Therefore,
there is no statistically significant evidence at α=0.01 to show that there is any change
in the mean ice thickness.
b.
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