Q1
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.
Q2
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
Q3
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.
Q4
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
Q5
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.
Q6
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)
Q7
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.
Q8
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.
Q9
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.