The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. The following year, a

Question

The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500. Is there sufficient evidence at the ?= 0.05 level to conclude that the mean cost has increased. Solve the question by traditional approach.

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Delwyn 3 years 2021-08-06T16:19:32+00:00 1 Answers 14 views 0

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  1. hongcuc2
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    2021-08-06T16:21:23+00:00
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    Answer:

    The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

    Step-by-step explanation:

    The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. Test if the mean cost has increased.

    At the null hypothesis, we test if the mean cost is still the same, that is:

    H_0: \mu = 13252

    At the alternative hypothesis, we test if the mean cost has increased, that is:

    H_1: \mu > 13252

    The test statistic is:

    We have the standard deviation for the sample, which means that the t-distribution is used to solve this question

    t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

    In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

    13252 is tested at the null hypothesis:

    This means that \mu = 13252

    The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500.

    This means that n = 20, X = 15560, s = 3500

    Value of the test statistic:

    t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

    t = \frac{15560 - 13252}{\frac{3500}{\sqrt{20}}}

    t = 2.95

    P-value of the test and decision:

    The p-value of the test is found using a t-score calculator, with a right-tailed test, with 20-1 = 19 degrees of freedom and t = 2.95. Thus, the p-value of the test is 0.0041.

    The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

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