Find the critical values and rejection regions for a rightta

Find the critical value(s) and rejection region(s) for a right-tailed chi-square test with a sample size n=15 and level of significance a = 0.05.

Round three decimal places.

A light bulb manufacturer guarantees that the mean of a certain type of light bulb is at least 761 hours. A random sample of 28 light bulbs has a mean of 736 hours. Assume the population is normally distributed and the population standard deviation is 61 hours. At significance level a =0.05, do you have enough evidence to reject the manufacturer

Solution

The degree of freedom =n-1=15-1=14

Given a=0.05, the critical value of chisquare with 0.95 and df=14 is 23.685 (from chisquare table)

The rejection region is if X^2 >23.685, we reject the null hypothesis.

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Let mu be the population mean

Null hypothesis: mu=761

Alternative hypothesis: mu<761 (i.e. claim)

It is a right-tailed test.

Given a=0.05, the critical value is Z(0.05) = -1.645 (from standard normal table)

The rejection region is if Z<-1.645, we reject the null hypothesis.

The test statistic is

Z=(xbar-mu)/(s/vn)

=(736-761)/(61/sqrt(28))

=-2.17

So we reject the null hypothesis

Yes, there is sufficient evidence to reject the claim that the mean bulb life is at least 761 hours