Question

A certain strain of bacteria occurs in all raw milk. Let X be the bacteria count per milliliter of milk and has a symmetrical mound shaped distribution. The mean of X is 2500 with a standard deviation is 301. In a large commercial dairy, the health inspector takes 49 random samples of milk produced every day. The sample mean of bacteria us x̄ was found to be 2605.

A. what is the distribution of x̄? Explain. What is the mean and standard deviation?

B. At 5% level of significance, the test hypothesis that cIaims the average bacteria count per ml of milk is greater than 2500. State Null and alternative hypothesis, critical value, test statistic. P-value and decision.

Answer 1

Part a)

Part b)

To Test :-

H0 :-  

H1 :-  

Test Statistic :-


Z = 2.4419

Test Criteria :-
Reject null hypothesis if


Result :- Reject null hypothesis

P value = P ( Z > 2.44 ) = 1 - P ( Z < 2.44) = 0.0073

Decision based on P value
P - value = P ( Z > 2.44 ) = 1 - P ( Z < 2.44) = 0.0073
Reject null hypothesis if P value < level of significance
P - value = 0.0073 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is sufficient evidence to support the claim that  the average bacteria count per ml of milk is greater than 2500 at 5% level of significance.