Question

A random sample has 49 values. The sample mean is 8.5 and the sample standard deviation is 1.5. Use a level of significance of 0.01 to conduct a left-tailed test of the claim that the population mean is 9.2. Compute the sample test statistic t. 0.005 0.0005 -2.267 -3.267

Answer 1

No,the population mean is not equal to 9.2 and the value in t statistic is -1.02.

Given sample size of 49,sample mean of 8.5,standard deviation of 1.5, significance level of 0.01.

We are required to find out whether the population mean is equal to 9.2 and the value of t in test statistic.

We have to first make the hypothesis for this.

$H_{0}$:μ≠9.2

$H_{1}$:μ=9.2

We have to use z statistic because the sample size is more than 30.

Z=(X-μ)/σ

We have been given sample mean but we require population mean in the formula so we will use sample mean.

Z=(8.5-9.2)/1.5

=-0.7/1.5

=-0.467

P value of -0.467 is 0.67975.

P value is greater than 0.01 so we will accept the hypothesis means population mean is not equal to 9.2.

t=(X-μ)/s/$\sqrt{n}$

=(8.5-9.2)/1.5/$\sqrt{49}$

=-0.7/0.21

=-1.02

Hence it is concluded that no,the population mean is not equal to 9.2 and the value in t statistic is -1.02.

Learn more about t test at https://brainly.com/question/6589776

#SPJ1