Question

The following table shows the average hourly wage rates or day-care centers from two locations based on two random samples. Use the table to complete parts a through c Location 1 Location 2 Sample mean Sample standard deviation Sample size $9.61 $1.28 27 $8.65 $1.25 34 a. Per or a hypothesis test using α =0.10 to determine e average hourly wage or day-care workers n Location 1 s $0.50 per hour higher than the average houry wage or day-care workers in Location 2 Assume the population variances for wage rates in each location are equal. Determine the null and alternative hypotheses for the test. Ho:H5-H2| |$0.50 H1 : μι_H2| |$0.50 Calculate the appropriate test statistic and interpret the result. The test statistic is Round to two decimal places as needed.) The critical value(s) is(are) Round to two decimal places as needed. Use a comma to separate answers as needed.) Because the test statistic the null hypothesis. b. Identify the p-value from part a and interpret the result The p-value is Round to three decimal places as needed.)

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Here we have T9.61,S, 1.28, n -27 -8.65,s2 -1.25, 2 34 Hypotheses are Test is right tailed. Level of significance: α =0.1 Since we can assume that variances are equal so degree of freedom of t test will be df=n,+n2-2-59 The pooled standard deviation is: (n, -1)s21+ (n-1)s -1.2633 Test Statistics ((9.61-8.65)-0.5)/(1.2633*sqrt((1/27)+(1/34))) n21.413 Critical value(CV):1.296 Rejection Region: If t> 1.296, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value: P-value-0.0815 Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value: ΞΤΙ NV(0.2,59) =TDIST(1413,59,1) Excel function for p-value:

(a)

The test statistics is

t = 1.41

The critical value is: 1.30

Since t lies in rejection region, we reject the null hypothesis.

(b)

The p-value is 0.0815

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