A study has a random sample of 34 subjects. The test statistic for testing Upper H 0 : mu equals 150 is tequals1.65. Find the approximate P-value for the alternative a. Upper H Subscript a Baseline : munot equals150, b. Upper H Subscript a Baseline : mugreater than150, and c. Upper H Subscript a Baseline : muless than150.

For the three different alternative hypotheses you have to calculate the corresponding p-value. The alternative hypothesis determines the "direction" of the hypothesis test, whether it is one-tailed or two tailed and in which tail you'll find the rejection region.

Remember, the p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

a)

H₁: μ ≠ 150

The rejection region is two-tailed, meaning that you'll reject the null hypothesis to small values of t and to high values of t, the p-value is also divided in two tails, one negative and one positive:

P(t₃₃≤-1.65) + P(t₃₃≥1.65)= P(t₃₃≤-1.65) + (1 - P(t₃₃≤1.65))= 0.0542 + (1 - 0.9458)= 2*0.0542= 0.1084

p-value= 0.1084

(Check first attachment for graph of the curves and p-value)

b)

H₁: μ > 150

The corresponding rejection region is one-tailed to the right, meaning that you'll reject the null hypothesis to high values of t, you'll find the p-value in the right tail of the distribution too and to calculate it you have to do as follows:

P(t₃₃≥1.65)= (1 - P(t₃₃≤1.65))= 1 - 0.9458= 0.0542

p-value= 0.0542

(See second attachment for graphic)

c)

H₁: μ < 150

This alternative hypothesis determines a rejection region one-tailed to the left, meaning that you'll reject the null hypothesis to small values of t.(see third attachment) The p-value will also be on the left (negative) tail of the distribution:

P(t₃₃≤-1.65)= 0.0542

p-value= 0.0542

I hope this helps!