Question

You wish to test the following claim ( H a ) at a significance level of α = 0.01 . H o : μ = 64.6 H a : μ < 64.6 You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n = 55 with mean ¯ x = 56.7 and a standard deviation of s = 15.9 . What is the test statistic for this sample? (Report answer accurate to two decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... less than (or equal to) α greater than α This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the population mean is less than 64.6. There is not sufficient evidence to warrant rejection of the claim that the population mean is less than 64.6. The sample data support the claim that the population mean is less than 64.6. There is not sufficient sample evidence to support the claim that the population mean is less than 64.6.

Answer 1

Here we are to test a one sample t test .

p-value=It is an evidence that how strong the sample data contradicts the null hypothesis,smaller the value of p-value greater the chance of rejecting the null.If the level of significance is given by alpha,then we reject the null hypothesis if the p-value is less than alpha.Using the p-value we can reach at a decision when the significance level alpha is known.

the p value of t test is nothing but the cdf of T  at a particular degrees of freedom.Here we calculate the value of the cdf or equivalently the p -value by using a online calculator for finding the cdf of t distribution.

here we are to put the value of df(54) and the calculated value of the test statistic T(which is -3.68) and just click on the button CALCULATE!