A social researcher wants to test the hypothesis that college students who drink

Question

A social researcher wants to test the hypothesis that college students who drink a lot while text messaging have more key strokes than those who do not drink while they text. The social researcher studies 50 drinking texters and 50 nondrinking texters. The average key strokes for the drinking texters was 142 with a standard deviation of 7.45. The average key strokes for the nondrinking texters was 134 with a standard deviation of 6.81.

1. What is the research hypothesis?

2. What is the null hypothesis?

3. Calculate t.

4. What are your degrees of freedom?

5. Are your results significant, and if so, at what level?

6. What are your conclusions about the null hypothesis?

Explanation / Answer

using minitab.

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 50 142.00 7.45 1.1
2 50 134.00 6.81 0.96

Difference = (1) - (2)
Estimate for difference: 8.00
95% lower bound for difference: 5.63
T-Test of difference = 0 (vs >): T-Value = 5.60 P-Value = 0.000 DF = 97

1&2

H0: drinking texters strokes < non drinking texters

Ha:drinking texters strokes > non drinking texters

3) T-Value = 5.60

4)df=97

5)significant at any level

because p=0

6) p < alpha

reject null hypothesis

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