Question

test the claim about the population mean μ at the level of significance α. assume the population is normally distributed.

claim: μ < 5015; α = 0.01 sample statistics: x̄ = 5117, s = 5725, n = 64

find the standardized test statistic t

t = 0.13
(round to two decimal places as needed.)
find the p - value

p = 0.551
(round to three decimal places as needed.)
decide whether to reject or fail to reject the null hypothesis. choose the correct answer below
reject h₀ there is enough evidence at the 1% level of significance to support the claim

Explanation:

Step1: Recall t - test formula

The formula for the standardized test statistic \( t \) in a one - sample t - test is \( t=\frac{\bar{x}-\mu}{s/\sqrt{n}} \), where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean under the null hypothesis, \( s \) is the sample standard deviation, and \( n \) is the sample size.
Given \( \bar{x} = 5117 \), \( \mu=5015 \), \( s = 5725 \), \( n = 64 \).
First, calculate the denominator \( s/\sqrt{n}=\frac{5725}{\sqrt{64}}=\frac{5725}{8}=715.625 \)
Then, calculate the numerator \( \bar{x}-\mu=5117 - 5015 = 102 \)
Now, calculate \( t=\frac{102}{715.625}\approx0.14 \) (Wait, the given t - statistic in the problem is 0.13, maybe due to rounding during intermediate steps. Let's check the P - value calculation. For a left - tailed test (since the claim is \( \mu<5015 \)), the degrees of freedom \( df=n - 1=64 - 1 = 63 \)
Using a t - distribution table or a calculator, for \( t = 0.13 \) and \( df = 63 \), the P - value for a left - tailed test is \( P(T<0.13) \). Since the t - statistic is positive and we have a left - tailed test, the P - value is \( 0.5 + P(0 < T<0.13) \). From t - distribution tables, for \( df = 60 \) (close to 63), \( P(0 < T<0.13)\approx0.051 \), so \( P - value\approx0.5+0.051 = 0.551 \) (which matches the given P - value).

Step2: Compare P - value and \( \alpha \)

The significance level \( \alpha=0.01 \). We know that if \( P - value>\alpha \), we fail to reject the null hypothesis \( H_0 \). Here, \( P - value = 0.551>0.01=\alpha \). So we fail to reject \( H_0 \). This means there is not enough evidence at the 1% level of significance to support the claim that \( \mu<5015 \).

Answer:

Fail to reject \( H_0 \). There is not enough evidence at the 1% level of significance to support the claim.