Q1athe critical z value is 205375if test value is gt 205375

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Reference no: EM13376441

a.The critical z value is 2.05375

If test value is > 2.05375 we reject null hypothesis.

b. test value= ( 47.1-45)/( 9/80^.5) = 2.087

c. p value = P( z > 2.087)= 0.018444

d. since p value < 0.02 we conclude that we reject null hypothesis.

a.The critical z value is 1.96

b. test value= ( .55-.45)/( .45*.55/300)^.5 = 1.1009

c. since test value < critical value we conclude that we do NOT reject null hypothesis.

d. p value = P( z > 1.1009)= 0.1354

a. Ho: p ≤ .55

H1 : p > .55

b. The critical z value is 1.645

sample proportion = 175/300= .583

test value= ( .583-.55)/( .45*.55/300)^.5 = 0.363

since test value < critical value we conclude that we do NOT reject null hypothesis.

The governor is NOT correct.

a. if null is true and we reject it this is a TYPE I error.

b. if null is not true and we accept it then TYPE II error is made

c. If null is true and we do not accept it then TYPE I error

d. IF null is not true and we accept it then TYPE II error

a. Ho: µ ≤ 85000

H1 : µ > 85000

b. The critical z value is 1.645

test value= (86200-85000)/(12000/80)^.5 = 0.894427

since test value < critical value we conclude that we do NOT reject null hypothesis.

The advertisement is NOT correct.

Assuming unequal variances.

Interval is ( difference in sample means) ± z value*SE

Se= standard error = (19²/ 24 +17.5²/28) ^.5 = 5.09

Difference in means= 130-125= 5

Interval is 5+/- 1.6*5.09= (-4.99, 14.99)

Ho: µ1 = µ2

H1; µ1 ≠ µ2

Test value is = ( difference in sample means) /SE

SE= (20.25²/ 36 +21.7²/31) ^.5 = 5.1556

Difference in means= 225-219= 6

Test value= 6/5.1556 = 1.163

Critical value is 1.645

As test value < critical value we do not reject the null hypothesis. There is no evidence that average speed was different in 2005 and 2006.

The test value is ( 12.45)/ (11/ √20) =5.06

The critical t value with 19 degrees is 2.43

As critical value > test value we do not reject null hypothesis. There is no evidence of difference in means.

Test value is = ( difference in sample proportions ) /SE

Sample proportion 1= 20/60=.5

Sample proportion2 = 24/80= 0.3

Difference = .2

SE= (.5*.5/60 +.3*.7/80) ^.5 = 0.082412

Test value= .2/.082412 = 2.4268

Critical value is 2.05375

As test value > critical value we reject the null hypothesis.

P value = P( z > 2.4268 ) = 0.007615

As p value is < 0.02 we reject null hypothesis.

Q10

Since this is a 2 tail test, we use the test value -1.547 and critical value = 2.94

Since the absolute value of test statistic < critical value we do not reject null hypothesis. There is no evidence of difference in price of campus and e retailer.

Reference no: EM13376441

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