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The sample mean and standard deviation from a random sample of 32 observations from a normal population were computed as x¯=26 and s=9. Test whether there is enough evidence to infer at the 10% significance level that the population mean is greater than 23. Find the critical region. (e.g. t > 1 for a one-tailed test, or t < -1 or t > 1 for a two-tailed test.)

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Answer:

There is enough evidence to infer at the 10% significance level that the population mean is greater than 23

Step-by-step explanation:

Given that the sample mean is 26 and sample std dev = 9

i.e

[tex]\bar x = 26\\s=9\\n = 32\\se = \frac{9}{\sqrt{32} } =1.591[/tex]

[tex]H_0: \bar x = 23\\H_a: \bar x >23[/tex]

(Right tailed test at 10% significance level)

Mean diff = 3

Test statistic t = [tex]\frac{3}{1.591} \\=1.886[/tex]

Critical region is t>1.310

df 31

p value = 0.034

Since p <0.10 reject null hypothesis

There is enough evidence to infer at the 10% significance level that the population mean is greater than 23

Critical region is in the enclosed file

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