Ten years ago, college students spent an average of 120 hours per semester on extra-curricular activities. A researcher believes that now c

Question

Ten years ago, college students spent an average of 120 hours per semester on extra-curricular activities. A researcher believes that now college students spend less time on extra-curricular activities than they did ten years ago. A simple random sample of 100 college students found that in the past year the average number of hours spent per semester in extracurricular activities was 107 hours with a standard deviation of 45 hours. If we are testing at a significance level of 0.05, based on the p-value, what is your conclusion?

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Ivy 1 week 2021-11-20T17:33:28+00:00 1 Answer 0 views 0

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  1. Emma
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    2021-11-20T17:35:12+00:00

    Answer:

    We conclude that the college students spend less time on extra-curricular activities than they did ten years ago.

    Step-by-step explanation:

    We are given that Ten years ago, college students spent an average of 120 hours per semester on extra-curricular activities.

    A simple random sample of 100 college students found that in the past year the average number of hours spent per semester in extracurricular activities was 107 hours with a standard deviation of 45 hours.

    Let \mu = average time spent on extra-curricular activities.

    So, Null Hypothesis, H_0 : \mu \geq 120 hours     {means that the college students spend more or equal time on extra-curricular activities than they did ten years ago}

    Alternate Hypothesis, H_A : \mu < 120 hours     {means that the college students spend less time on extra-curricular activities than they did ten years ago}

    The test statistics that would be used here One-sample t test statistics as we don’t know about the population standard deviation;

                           T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

    where, \bar X = sample average number of hours spent per semester = 107 hrs

                 s = sample standard deviation = 45 hours

                 n = sample of college students = 100

    So, test statistics  =  \frac{107-120}{\frac{45}{\sqrt{100} } }  ~  t_9_9

                                   =  -2.889

    The value of t test statistics is -2.889.

    Now, the P-value of the test statistics is given by following formula;

                   P-value = P( t_9_9 < -2.889) = 0.00314

    Since, the P-value is less than the level of significance as 0.05 > 0.00314, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

    Therefore, we conclude that the college students spend less time on extra-curricular activities than they did ten years ago.

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