Question

A sample of 36 observations is selected from a normal population. The sample mean is 49, and the population standard deviation is 5. Conduct the following test of hypothesis using the 0.05 significance level.

H₀: μ = 50

H₁: μ ≠ 50

1. Is this a one- or two-tailed test?

• One-tailed test

• Two-tailed test done

2. What is the decision rule?

• Reject H₀ if −1.96 < z < 1.96

• Reject H₀ if z < −1.96 orz > 1.96 done

3. What is the value of the test statistic? (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

Ans : -1.2. done

4. What is your decision regarding H₀?

• Reject H₀

• Fail to reject H_{0 done}

1. e-1. What is the p-value?

0.2301

1. e-2. Interpret the p-value?

Ans. : 23.01% wrong

Answer 1

a) Null hypothesis Ho: population mean is so Ho ll =50 50 Alternative hypothesis H, i population mean is not equal to Hi, ll + 50 (too taited This is two tailed tot Decision a) rule. f& two tailed tyt The critical Calue at x=0.05 is + 1.96 Reject to it <-1096 & Z>2.96.

el a) The tept Statistic all ~N 2011) 2 lus -= = 5, Given that 49, x = n=36, el=50 under Ho, the tege Statistic is Z ~NLOil - ell न 49-50 36 ilole oll 5 6 = -lo 2 =-12 Cal Z tab fail to a 71.96 - 1096 cecal <1096 <1096 rject to e-) Pialue P = P{ z = 1213 teoo tooted - P{ze 1.23 x=0.05 = 0.230139 foomnende table 0.2301

al Intepretation The result is not significant at P <0.05 P value is 0.3 >0.05 we fail to reject Ho.