Question

While there is some expected variation in quantities

provided by the quarry​ (the standard deviation is known to be 15.34

15.34 ​tons/week, regardless of the mean​ supplied), they require 83

83 ​tons/week, on average. If it can be proven that the supplier is not meeting that​ requirment, then they will find a second quarry to source from. For 36

36 ​weeks, the mean supply was 75.7

75.7 ​tons/week. Is there sufficient evidence to conclude that they need a second supplier.

What are the null and alternative hypotheses for the appropriate​ test?

Runs the test at a 1​% significance​ level, using each of the three equivalent methods

The sample mean is:

The critical value for the sample mean is:

Since the sample mean is greater or less than the critical value​ (PE) do not reject or reject the null hypothesis and conclude that there is

sufficient or insufficient evidence to prove that the population mean suppy is less than 83 tons/week.

The test statistic is

The critical value for the test static is

Since the test statistic is greater or less than the critical value​ (TS), do not reject or reject the null hypothesis and conclude that there is sufficient or insufficient evidence to prove that the population mean supply is less than 83tons/week.

The​ p-value for this test is

Since the​ p-value is greater or less than the significance​ level,α ​do not reject or reject the null hypothesis and conclude that there is sufficient or insufficient evidence to prove that the population mean supply is less than 83tons/week.

What is the appropriate action given the evidence​ above?

Do not look for another or Find a second quarry

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