Respuesta :
Answer:
a. Two - tailed test
b. Decision rule is that since hypothesis is of lower tail, so we will reject null hypothesis if the test statistics is smaller than the stated critical value
c. Value of test statistics = -2.667
d. We will reject [tex]H_0[/tex] .
e. P-value = 3.85 x [tex]10^{-3}[/tex]
Step-by-step explanation:
We are given that ;
Null Hypothesis, [tex]H_0[/tex] : μ ≥ 220
Alternate Hypothesis, [tex]H_1[/tex] : μ < 220
a. This is one - tail test as in alternate hypothesis we are only concerned with value of [tex]\mu[/tex] less than 220 not of less than and greater than both i.e. there is no [tex]\neq[/tex] sign in [tex]H_1[/tex] .
b. Decision rule is that since hypothesis is of lower tail, so we will reject null hypothesis if the test statistics is smaller than the stated critical value .
c. Test Statistics is given by;
[tex]\frac{xbar - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] follows N(0,1) where, xbar = sample mean = 215
[tex]\sigma[/tex] = Population standard deviation = 15
n = sample size = 64
⇒ [tex]\frac{215 - 220}{\frac{15}{\sqrt{64} } }[/tex] = -2.667
So, the value of test statistics = -2.667 .
d. At 0.025 level of significance, the z table gives the critical value of -1.96.
Since our test statistics is less than the critical value as -2.667 < -1.96 so we will reject null hypothesis [tex]H_0[/tex] .
e. P-value is given by, P(Z< -2.667) = P(Z > 2.667) = 0.00385 = 3.85 x [tex]10^{-3}[/tex] .
