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.

Answer 1

The claim is false that the population means 9.2. and the required value of the t statistic is -1.02.

Given that,
Sample size (n) = 49,
Sample means (μ) = 8.5,
Standard deviation (σ) = 1.5, and the significance level of 0.01.
To determine whether the population means is equal to 9.2 and also the value of sample test statistic t.

What is T- statistic?

The T - statistic is the ratio of the predictive value of a parameter from its hypothesized value to its standard disputable values.

Let to prove the claim let's assume the population means is not equal to 9.2.
Now

Z = (x - μ)/σ

Z = (8.5 - 9.2) / 1.5

  = -0.7 / 1.5

  = -0.467

The probability(P) of z of -0.467 is 0.67975.

Since the value of P is more than 0.01 so it implies that the assumption that population means is not equal to 9.2.

$t = \frac{x - \mu}{\sigma/\sqrt{n} }\\ t = \frac{8.5 - 9.2}{1.5/\sqrt{49} } \\ t = -1.02$

   

Thus, the claim is false that the population means 9.2. and the required value of the t statistic is -1.02.

Learn more about T-statistic here:
brainly.com/question/15236063

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