Question

1. The national mean annual salary for school administrator is $90,000 a year (The Cincinnati Enquirer, April 7, 2012). A school official took a sample of 25 school administrators in the state of Ohio to learn about salaries in that state to see if they differed from the national average.

1. Formulate hypotheses that can be used to determine whether the population mean annual administrator salary in Ohio differs from the national mean of $90,000.
2. The sample data for 25 Ohio administrators is contained in the file named Administrator. What is the p-value for the hypothesis test in part (a)?
3. At α = 0.05, can your null hypothesis be rejected? What is your conclusion?
4. Repeat the preceding hypothesis test using the critical value approach.
   Insert Page Layout Formulas Data Review View Home 3bCalibri (Body) 11AA Paste G4 1 Salary 77600 76000 90700 97200 90700 7101800 78700 81300 10 84200 11 97600 12 77500 13 75700 14 89400 15 84300 16 78700 17 84600 87700 19 103400 20 83800 21101300 22 94700 23 69200 24 95400 25 61500 26 68800 8 18 27 28 29
   

Answer 1

Date: 14/05/2019 Answer To test the hypothesis is that the population mean annual administrator salary in Ohio differs from the national mean of $90,000 at 5% significance level. a) The null and alternative hypothesis is, H H 90000 На : # $90000 b) The t-test statistics is, By using MINITAB, find t-test statistics with the help of following steps is 1) Import the data. 2) Select the Stat and choose the Basic Statistics option 3) Select the 1 sample t and choose variable option and put Sample in column 4) Give the Hypothesized mean 5) Click option button choose level of significance and alternative hypothesis. 6) Click Ok.

One-Sample T: Salary Test of mu90000 vs not90000 Variable N Mean StDev SE Mean 95% CI Salary 25 85272.0 11039.2 2207.8 (80715.2, 89828.8) 2.14 0.043 From the MINITAB output, thet-test statistics is-2.14 Thep-value for this test is From the MINITAB output, thep-value for this test is 0.043 c) Decision The conclusion is that the p-value in this context is less than 0.05 which is 0.043, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the population mean annual administrator salary in Ohio differs from the national mean of S90,000. The result is statistically significant

d) Critical value approach differs from the national mean of $90,000 at 5% significance level The null and alternative hypothesis is, Ho: μ-390000 Ha : μ # $90000 The t-test statistics is, First, compute mean and standard deviation then find the t-test statistics he sample mean is, 1l 77600 76000 +.61500 68800 25 = 85272

The sample standard deviation is, n-1 (77600 - 85272)(76000 - 85272)2 + ...+ (61500 85272)2 (68800 -85272) 25-1 = v12 1864600 = 1 1039.23 The t-test statistics is, 85272 90000 11039 23/25 =-2.14 The t-test statistics is-2.14

The t critical value is, The sample size is small and two-tailed test. Look in the column headed α = 0.05 and the row headed in the t distribution table by using degree of freedom is, d.f.= n-1 = 25-1 = 24 The t critical value is 2.064 Decision The conclusion is that the t value corresponds to sample statistics is falls in the critical region, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the population mean annual administrator salary in Ohio differs from the national mean of S90,000. The result is statistically significant.