Question

1. In a simple random sample of n = 840 voters, X = 609 indicate that they will start holiday shopping this weekend. (a) Find the 95% confidence interval for p and the margin of error. (b) Use the six-step method of hypothesis testing to determine if the proportion is significantly different from 0.75. Use α = 0.05. (c) Is your conclusion the same with the test and the confidence interval? (d) How would the p-value change if you used the alternative hypothesis HA : p < 0.75. Would this change your conclusion?

2. A random sample of n = 10 statistics classes is selected. The number of students in the 10 classes is: 25 28 32 33 37 39 42 44 46 47 (a) Find the mean, standard deviation, and five-number summary. Use the 1.5 × IQR rule to determine if there are any outliers in the data. (b) Find the 90% CI for µ (c) Use the one-sample t-test to test the claim that the mean is greater than 32. Use the six-step hypothesis testing method and α = 0.05.

3. Most modern email clients have spam filtering, or methods to try to prevent unwanted junk email (spam) from being delivered. Many of these filters are based on statistical methods. Suppose we define the null hypothesis, HO, to be that a piece of email is okay and should be delivered to your Inbox, and the alternative hypothesis, HA, to be that a piece of email is spam and should be delivered to the Trash folder. (a) What is “rejecting the null”, in the context of email? (b) What is “failing to reject the null”, again in context? (c) What would Type I and Type II error be in this situation? (d) Suppose the system administrator increased the “power” of the email filter by decreasing the number of points it takes for an email to be rejected and sent to the Trash folder. This is similar to increasing α in a statistical test. Why might this be a bad idea?

Answer 1

Question 1:

(a) Find the 95% confidence interval for p and the margin of error.

The 95% confidence interval for p is between 0.6948 and 0.7552 and the margin of error is 0.0302.

(b) Use the six-step method of hypothesis testing to determine if the proportion is significantly different from 0.75. Use α = 0.05.

The hypothesis being tested is:

H0: p = 0.75

Ha: p ≠ 0.75

Observed	Hypothesized
0.725	0.75	p (as decimal)
609/840	630/840	p (as fraction)
609.	630.	X
840	840	n
	0.0149	std. error
	-1.67	z
	.0943	p-value (two-tailed)

The p-value is 0.0943.

Since the p-value (0.0943) is greater than the significance level (0.05), we fail to reject the null hypothesis.

Therefore, we cannot conclude that the proportion is significantly different from 0.75.

(c) Is your conclusion the same with the test and the confidence interval?

Yes

(d) How would the p-value change if you used the alternative hypothesis HA : p < 0.75. Would this change your conclusion?

The p-value is 0.0471.

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Thank you! :)