The USDA reports that 25 of fourmember households spend less

The USDA reports that 25% of four-member households spend less than $100 per week on groceries. Suppose a random sample of 500 four-member households is taken. Find the probability that greater than 111 households spend less than $100 per week on groceries. There is a 20% probability that the sample proportion of households who spend le\\s han $100 per week on groceries is no less than find the probability that the sample proportion of households who spend less than SI00 per week on groceries is between 24% and 30%. In 2009, the average starting salary for an individual right out of college with a bachelor\'s degree in business w as $49,400 per year. Suppose a 2014 study, which randomly sampled 48 individuals right out of college with a bachelor\'s degree in business, showed an average starring salary of $51,620. Assume the population standard deviation is $7,810. Does the 2014 study suggest that the asenigc starting salary of an individual right out of college with a bachelor\'s degree in business is now higher at alpha =.025? What is the alternative hypothesis? What is the decision rule? hat is the test statistic? What is the conclusion?. W hat is the p-value? hat is the 80% confidence interval

Solution

C

Let X_i be the status of i^th household. That means X_i be 1 if the i^th household spends less than $100 per week, and 0 otherwise.

Clearly, X_i ~Bernoulli(p) , for i=1,2,..,500 & where p=0.25 (from survey report)

Let X be the the sum of all X_is for i=1,2,..,500

Then X~BInomial(500 , 0.25)

E(X)=500*0.25=125

Var(X)=500*0.25*(1-0.25)=500*0.25*0.75=93.75

Now, apply normal approximation as

X~N(125 , 93.75)

(1)

P(X>111)=0.92

(2)

0.8^th quantile of X is: 133.149

So, P(X>133) =0.2 (approx)

Hence, 133 is the required answer.

(3)

24% of 500=120

30% of 500=150

P(120<=X<=150)=0.7