Question

Consider the following hypotheses: HO: > 220 HA: U <220 A sample of 72 observations results in a sample mean of 209. The population standard deviation is known to be 18. (You may find it useful to reference the appropriate table: z table or t table) a-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Test statistic ſ -5.19 a-2. Find the p-value. 0.05 s p-value < 0.10 p-value = 0.10 p-value < 0.01 0.01 s p-value < 0.025 0.025 s p-value < 0.05 b. Does the above sample evidence enable us to reject the null hypothesis at a = 0.10? Yes since the p-value is less than the significance level. No since the p-value is greater than the significance level. No since the p-value is less than the significance level. Yes since the p-value is greater than the significance level.

Answer 1

Solution :

$\mu$= 220

$\bar x$ =209

$\sigma$ =18

n = 72

This is the left  tailed test .

The null and alternative hypothesis is

H0 :  $\mu$ ≥220

Ha : $\mu$ < 220

a1) Test statistic = z

= ($\bar x$ - $\mu$ ) / $\sigma$ / $\sqrt$ n

= (209 -220) / 18 / $\sqrt$ 72

= -5.185 = -5.19

2) P (Z < -5.19) =0

P-value = 0.0000

p-value < 0.01

b) $\alpha$ = 0.10

0 < 0.10

Yes since the p value less than the significance level.

c) $\alpha$ = 0.01

Yes since the p value less than the significance level.

d) $\alpha$ = 0.01

We conclude that the population mean is less than 200.