Question

For a random sample of 36 data pairs, the sample mean of the differences was 0.72. The sample standard deviation of the differences was 2. At the 5% level of significance, test the claim that the population mean of the differences is different from 0. (a) Is it appropriate to use a Student's t distribution for the sample test statistic? Explain. No, the sample size is not larger than 30. Yes, the sample size is larger than 30. No, the standard deviation is not smaller than the sample mean. Yes, the standard deviation is larger than the sample mean. What degrees of freedom are used? (b) State the hypotheses. OHO: 40 = 0; M:49+0 Ho: Mg = 0; Mq:47 > 0 Hoi flg = 0; H: 490 Hoillg+ 0; H: 4, = 0 (c) Compute the t value. (Round your answer to three decimal places.) (d) Estimate the P-value of the sample test statistic. OP-value > 0.500 0.250 < P-value < 0.500 O 0.100 < P-value < 0.250 O 0.050 < P-value < 0.100 0.010 < P-value < 0.050 OP-value < 0.010

Answer 1

Answer Date: 11/11/2019 It is appropriate to use a Student'st distribution for the sample statistics due to the sample size is higher than 30. The answer option is 2). The degrees of freedom are, d.f.=n-1 = 36-1 = 35 The degrees of freedom are 35). b) To test the hypothesis is that the population mean of difference is different from 0 at 5% significance level.

The null and alternative hypothesis is, H:lq = 0 H:lla 70 The answer option is 1. The t-test statistics is, sin 0.72 (2)/36 = 2.160 The t-test statistics is 2.160.

The p-value for this test is, From the t distribution table, show the degree of freedom 35 and t-test statistics is 2.160 and the p-value is between 0.010 and 0.050. The answer option is 5. e) Decision The conclusion is that p-value in this context is less than 0.05, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the population mean of difference is different from 0. The result is statistically significant. The answer option is 2). To test the hypothesis is that the population mean change increase in corporate revenue is different from the population mean percentage increase in CEO salary.

The level of significance is 5%. The null and alternative hypothesis is, Hola = 0 H:1000 The answer option is 4. The Student's t sampling distribution will use due to the difference (d) has an approximately normally distributed. The answer option is 2). The t-test statistics is, By using Excel software, find t-test statistics with the help of following steps: 1) Import the data. 2) Select t-test: Paired Two Samples for Means from Data Analysis in Data menu.

4) 5) Select Variable 1 Range and Variable 2 Range. Give Hypothesized Mean Difference. Select Labels and give Alpha. Choose Output Range from Output options. Click Ok. t-Test: Paired Two Sample for Means Before After 20.25 17.5 118.7857143 100.857143 Mean Variance Observations Pearson Correlation Hypothesized Mean Dit 0.740030436 df t Stat 1.024483428 From the Excel output, the t-test statistics is 1.024].

The p-value is, 1) By using Imathas hypothesis test graph generator, find p-value with the help of following steps: Select t distribution and give the Sample size. Select the Left tailed type. Give Test Statistics Value. Select Shade P-value region. Click Draw button. P-Value=0.3397 From the Imath as hypothesis test graph generator, the p-value is 0.3397) between 0.250 and 0.500.

d) Decision The conclusion is that the significant p-value in this context is higher than 0.05 which is less than 0.3399, so the null hypothesis is fail to reject at 5% level of significance. The result is not statistically significant. The answer option is 1). e) There is in sufficient evidence to indicate that the population mean change increase in corporate revenue is different from the population mean percentage increase in CEO salary. The answer option is 3.