Question

Every Easter, an average of 960,000 hams and 90,000 bottles of wine are sold per state. Importantly, the population variance/standard deviation is not known. Below is the number of hams and bottles of wine sold in each of the ten northeast states.

State	# of Hams	# Bottles of Wine
Maine	199,700	45,900
Vermont	93,800	25,500
New Hampshire	200,200	60,000
Massachusetts	1,020,100	294,250
Rhode Island	200,000	74,000
Connecticut	525,100	111,250
New York	2,962,500	780,600
Pennsylvania	1,920,200	480,000
New Jersey	1,341,600	355,400
Delaware	142,500	105,500

Use the above data to answer the following question (each question is worth 1.5 pts):

What is the mean and estimated population variance for both the number of hams and number of bottles of wine sold in the American northeast?

If we are interested in whether or not the average number of hams sold in the American northeast differs from that of the rest of the American population, how would we write the null and alternative hypotheses (using the new statistical notation for stating statistical hypotheses; see page 235)?

If we are interested in whether or not the average number of bottles of wine sold in the American northeast is greater than that of the rest of the American population, how would we write the null and alternative hypotheses (using the new statistical notation for stating statistical hypotheses; see page 235)?

What is t-crit and t-obt for the hypothesis test stated in question #2? Is the hypothesis test significant or insignificant? Would you reject or fail to reject the null hypothesis assuming an alpha-level of 0.05?

What is t-crit and t-obt for the hypothesis test stated in question #3? Is the hypothesis test significant or insignificant? Would you reject or fail to reject the null hypothesis assuming an alpha-level of 0.05?

Answer 1

2) NULL HYPOTHESIS H0:

AVERAGE HYPOTHESIS Ha:

ALPHA= 0.05

t= (860570-960000)/961656.929/sqrt(10)

t= -0.3264

TWO TAILED TEST

Degrees of freedom= 10-1=9

t crit= 2.26

Since calculated value t is less than critical value of t hence hypothesis test is insignificant. Fail to reject null hypothesis H0.

3) NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS HA:

ALPHA= 0.05

t= 233240-90000/245845.3622/sqrt(10)

t= 1.84

degrees of freedom= 10-1=9

t critical= 1.83

Since critical t value is smaller than calclated t value therefore we reject null hypothesis .Test is significant.