Question

a company manufactures tennis balls. when its tennis balls are dropped onto a concrete surface from a height of 100 inches, the company wants the mean height the balls bounce upward to be 54.9 inches. this average is maintained by periodically testing random samples of 25 tennis balls. if the t - value falls between $-t_{0.95}$ and $t_{0.95}$, then the company will be satisfied that it is manufacturing acceptable tennis balls. a sample of 25 balls is randomly selected and tested. the mean bounce height of the sample is 56.5 inches and the standard deviation is 0.25 inch. assume the bounce heights are approximately normally distributed. is the company making acceptable tennis balls? find $-t_{0.95}$ and $t_{0.95}$. $-t_{0.95}=\square$ $t_{0.95}=\square$ (round to three decimal places as needed.)

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\), where \(n = 25\). So \(df=25-1 = 24\).

Step2: Find \(t_{0.95}\) using the t - distribution table

Looking up the value in the t - distribution table (or using a calculator with a t - distribution function) for a two - tailed test with \(\alpha=1 - 0.95=0.05\) and \(df = 24\).
The value of \(t_{0.95}\) is such that the area to the right of \(t_{0.95}\) is \(\frac{1 - 0.95}{2}=0.025\) and the area to the left of \(-t_{0.95}\) is also \(0.025\).
From the t - distribution table or using a calculator (e.g., in R: qt(0.975,24)), we get \(t_{0.95}=2.064\).
Since the t - distribution is symmetric about \(t = 0\), \(-t_{0.95}=- 2.064\).

Answer:

\(-t_{0.95}=-2.064\)
\(t_{0.95}=2.064\)