Question

1- Students had a mean grade of 70% in the past. After applying a new teaching method, the following scores were recorded for a sample of 9 random students:

Scores
73
63
64
61
66
69
61
70
79

In order to construct a 95% confidence interval for the new mean grade, we should use:

• 1-PropZInt
• ZInterval
• 2-SampZInt
• 2-SampTInt
• TInterval
• 2-PropZInt

2-

Assume that a sample is used to estimate a population proportion p. Find the 99.5% confidence interval for a sample of size 336 with 81 successes. Enter your answer using decimals (not percents) accurate to three decimal places.

< p <

3-

A test was given to a group of students. The grades and gender are summarized below:

	A	B	C	Total
Male	17	4	10	31
Female	2	9	16	27
Total	19	13	26	58

Let pp represent the percentage of all female students who would receive a grade of A on this test. Use a 80% confidence interval to estimate pp to three decimal places.

Enter your answer using decimals (not percents).

<p <

4-

Out of 500 people sampled, 295 preferred Candidate A.

Based on this, find a 90% confidence level for the true proportion of the voting population (pp) prefers Candidate A.

Give your answers as decimals, to three places.

<p<

Answer 1

1:

Since population standard deviation is unknown so t interval should be used. Correct option is:

TInterval

2:

3:

Out of 27 females, 2 receive A grade. So,

4: