From her firm’s computer telephone log, an executive found that the mean length of 68 telephone calls during July was 4.38 minutes with a st

Question

From her firm’s computer telephone log, an executive found that the mean length of 68 telephone calls during July was 4.38 minutes with a standard deviation of 5.52 minutes. She vowed to make an effort to reduce the length of calls. The August phone log showed 44 telephone calls whose mean was 2.458 minutes with a standard deviation of 2.506 minutes. Steps: 1. Open Minitab on your computer. 2. There is no data file for this example. The information above contains summarized data (take info from story and plug into Minitab) (a) Choose the appropriate hypotheses for a right-tailed test to see if calls were quicker, on average, in August. Assume µ1 is the average call length in July and µ2 is the average call length in August. a. H0: μ1 – μ2 ≤ 0 vs. H1: μ1 – μ2 > 0 b. H0: μ1 – μ2 < 0 vs. H1: μ1 – μ2 ≤ 0 a b (b-1) Obtain a test statistic tcalc and p-value assuming unequal variances. (Use Minitab. Round your answers to 3 decimal places.) tcalc p-value (b-2) Interpret the results using α = .01. the null hypothesis.

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Ivy 4 months 2021-10-08T15:47:55+00:00 1 Answer 0 views 0

Answers ( )

  1. Emma
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    2021-10-08T15:49:41+00:00

    Answer:

    Test Statistics = 2.167

    P-value = 1.76%

    Conclusion : We will accept null hypothesis.

    Step-by-step explanation:

    We are given that an executive found that the mean length of 68 telephone calls during July was 4.38 minutes with a standard deviation of 5.52 minutes. She vowed to make an effort to reduce the length of calls. The August phone log showed 44 telephone calls whose mean was 2.458 minutes with a standard deviation of 2.506 minutes.

    Let \mu_1 = average call length in July  and  \mu_2 = average call length in August

    Null Hypothesis, H_0 : \mu_1 - \mu_2 <= 0      or   \mu_1 <= \mu_2  

    Alternate Hypothesis, H_1 : \mu_1 - \mu_2 > 0   or  \mu_1 > \mu_2

    Now, the test statistics used here will be;

             T.S. = \frac{(X_1bar - X_2bar)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~ t_n__1+n_2-2

    where, X_1bar = mean length of 68 telephone calls during July =4.38 minutes                      X_2bar = mean length of 44 telephone calls during August = 2.458 minutes

      n_1 = number of telephone calls in July = 68

      n_2 = number of telephone calls in August = 44

       s_p = \sqrt{\frac{(n_1-1)s_1^{2}_(n_2-1)s_2^{2}  }{n_1+n_2-2} } = \sqrt{\frac{(68-1)5.52^{2}+(44-1)2.506^{2}  }{68+44-2} } = 4.584

    So, now test statistics = \frac{(4.38 - 2.458)-0}{4.584\sqrt{\frac{1}{68} +\frac{1}{44} } } = 2.167

    Now, at 1% level of significance, the t table gives critical value of 2.358 at 110 degree of freedom. Since our test statistics is less than the critical value so we have insufficient evidence to reject null hypothesis and conclude that the mean length of calls are also same for August as that of July.\

    P-value is given by = P(t_1_1_0 > 2.167) = 0.01758 or 1.76% {using t table}

    Here, also P-value is greater than significance level as 1.76% > 1% so we will accept null hypothesis.

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45:7+7-4:2-5:5*4+35:2 =? ( )