For this question, I'm not sure how I would solve for the test statistic and the p-value calculation after it? When it is operating correctly, a machine for manufacturing tennis balls produces balls with a mean weight of 57.7 grams. A random sample of eight balls from the last batch produced has the weights listedbelow. Use the Inference for a Mean web app to complete parts a through c. Be sure to use the data in this exercise and not the preloaded data when using the web app.57.6 57.4 57.7 57.6 57.6 57.3 57.957.3a. Find the test statistic and P-value for a test of whether the process is in control against the alternative that the true mean of the process now differs from 57.7.The test statistic is 0.(Round to two decimal places as needed.)

The following data has been provided: 57.6 , 57.4 , 57.7, 57.6 ,57.6, 57.3, 57.9, 57.3 The sample size is n=8. The hypothesized population mean is μ=57.7. We will use the t test for one population mean because the population standard deviation is unknown and the sample size is less than 30. Firstly we need to compute sample mean and sample standard deviation. x x2 57.6 3317.76 57.4 3294.76 57.7 3329.29 57.6 3317.76 57.6 3317.76 57.3 3283.29 57.9 3352.41 57.3 3283.29 ∑x=460.4 ∑x2=26496.32 Mean: x¯=∑x n =57.6+57.4+57.7+57.6+57.6+57.3+57.9+57.38 =460.48 =57.55Standard Deviation: S=∑x2-∑x 2n n-1 =26496.32-460.4288-1 =26496.32-26496.027 =0.207 Null and Alternative Hypothesis: H0:μ=57.7H1:μ≠57.7 Test Statistic: t=x¯-μSn =57.55-57.70.2078 =-0.150.073 =-2.05 The test statistic is t=2.05. The degrees of freedom are : df=n-1 =8-1 =7 The p-value can be computed using the EXCEL formula: p-value=T.DIST.2T(t,df) =T.DIST.2T(2.05,7) =0.08 The level of significance is α=0.05. Decision Rule: We reject the null hypothesis if the p-value is less than the level of significance. Decision: We fail to reject the null hypothesis because p-value=0.08>α=0.05. Conclusion: We fail to reject the null hypothesis. There is not enough evidence to conclude that the true mean differs from 57.7.