Question

The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar businesses reported the mean number of customers waiting to have their oil changed on Saturday morning is 3.6. Suppose the local oil changing business owner, wants to perform a hypothesis test. The null hypothesis is the population mean is 3.6 and the alternative hypothesis is the population mean is not equal to 3.6. The owner takes a random sample of 16 Saturday mornings during the past year and determines the sample mean is 4.2 and the sample standard deviation is 1.4. It can be assumed that the population is normally distributed. The level of significance is 0.05. The decision rule for this problem is to reject the null hypothesis if the observed "t" value is _______.

		less than -2.131 or greater than 2.131
		less than -1.761 or greater than 1.761
		less than -1.753 or greater than 1.753
		less than -2.120 or greater than 2.120
		less than -3.120 or greater than 3.120

Answer 1

We have

Ho: mu = 3.6

Hq:mu≠3.6

sample mean(x bar)=4.2

Sample standard deviation (s)=1.4

Sample size (n)=16

Test statistic t= (xbar-mu)/(s/√n)

t=(4.2-3.6)/(1.4/√16)=1.7142

Observed test statistic (t)= 1.714

Df=(n-1)=15

Level of significance==0.05

Critical value(at alpha/2=0.025 and 15 df)=±2.1314

We Reject the null hypothesis if the observed t value is less than -2.131 or greater than 2.131

Ans→less than -2.131 or greater than 2.131

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{Extra→ we fail to reject Ho here as t lies in this interval}