Question

please answer all 4♥️

Find the critical value(s) for a left-tailed z-test with a = 0.06. Include a graph with your answer. The critical value(s) is(are) (Round to two decimal places as needed. Use a comma to separate answers as needed.)

Answer 1

-1.55 Critical value: Rejection Region: If z<-1.55, Reject HO

Excel function for critical value: =NORMSINV(0.06)

Following is the graph:

0 درنه

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Here we have x=1281.95, =205.18, n =250 Hypotheses are: H: US 1250 H: > 1250 Level of significance: a=0.03 Test is one tailed (right tailed). Test Statistics: Z -u olun = (1281.95-1250)/(205.18/sqrt(250)) = 2.46 Critical value: Rejection Region: 1.881 If z > 1.881, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis. P-value: P-value= 1 - Plz<=2.46) = 0.007 Decision: Since p-value is less than level of significance so we reject the null hypothesis. Excel function for critical value: Excel function for p-value: =NORMSINV(1-0.03) =1-NORMSDIST(2.46)

Answer: z = 2.46

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Answer: z = 2.48

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Here we have 3 =269,0 =36, n=81 Hypotheses are: H:us 260 H:11 > 260 Level of significance: a =0.03 Test is one tailed (right tailed). Test Statistics: Z= T-ul (269-260)/(36/sqrt(81)) ola = 2.25 Critical value: Rejection Region: 1.881 If z > 1.881, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis. P-value: P-value= 1 - Plz<=2.25) = 0.012 Decision: Since p-value is less than level of significance so we reject the null hypothesis. Excel function for critical value: Excel function for p-value: =NORMSINV(1-0.03) =1-NORMSDIST(2.25)

Answer: z = 2.25