Question

In: Statistics and Probability

Ramilda wants to see if other students in her class listen to podcasts for a longer...

  1. Ramilda wants to see if other students in her class listen to podcasts for a longer amount of time than students across the country. She collects data from twenty students in her class concerning the number of minutes per day they listen to podcasts. On average people in the United States listen to podcasts for 93 minutes per day. The data from her participants are summarized below:

Participant

Minutes

1

150

2

45

3

67

4

221

5

30

6

109

7

78

8

126

9

135

10

58

11

89

12

90

13

93

14

108

15

119

16

185

17

100

18

109

19

189

20

167

  1. What are the null and alternative hypotheses using mathematical notation (1 point)?
  1. Compute the appropriate test statistic for testing the hypothesis (5 points).

  1. Using  = .05, what do you conclude about how Remilda’s class compares to other students across the country in the amount of time they spend listening to podcasts? Be sure to provide the critical value in your answer (1.5 points).

Solutions

Expert Solution

Given: = 93, s = 113.4, = 49.84 , n = 20, = 0.05

(a) The Hypothesis:

H0: = 93

Ha: > 93

This is a right tailed Test.

(b) The Test Statistic: Since the population standard deviation is unknown, we use the students t test.

The test statistic is given by the equation:


(c) The Critical Value:   The critical value (Right Tail) at = 0.05, for df = 19, tcritical = +1.691

The p Value:    The p value (Right Tail) for t = 2.42, for degrees of freedom (df) = n-1 = 19, is; p value = 0.0105

The Decision Rule:   If tobserved is > tcritical, then Reject H0

Also if P value is < , Then Reject H0.

The Conclusion:

Since tobserved (2.44) is > tcritical (1.691), We Reject H0. There is sufficient evidence at the 95% significance level to conclude that the mean time spent on listening to podcasts is greater than 93.

___________________________________________________

Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)2.

Participant Minutes Mean (X - Mean)2
1 150 113.4 1339.56
2 45 113.4 4678.56
3 67 113.4 2152.96
4 221 113.4 11577.76
5 30 113.4 6955.56
6 109 113.4 19.36
7 78 113.4 1253.16
8 126 113.4 158.76
9 135 113.4 466.56
10 58 113.4 3069.16
11 89 113.4 595.36
12 90 113.4 547.56
13 93 113.4 416.16
14 108 113.4 29.16
15 119 113.4 31.36
16 185 113.4 5126.56
17 100 113.4 179.56
18 109 113.4 19.36
19 189 113.4 5715.36
20 167 113.4 2872.96
Total 2268 Total 47204.8
Mean 113.4 Variance 2484.46316
SD 49.84

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