Statistics help please?? null hypothesis?? alternative hypothesis?

Question

Our company's new "Lite" yogurt is supposed to contain only 1.5 grams of fat. The FDA monitors these advertising claims for accuracy. If they can prove, at the 10% significance level that our product contains more fat than advertised, then we will be subject to fines and product recall. A sample of 90 packages yielded an average of 1.54 grams of fat, with a standard deviation of 0.16. Perform the appropriate hypothesis test to see if our product violates FDA advertising standards.

Write the appropriate null and alternative hypotheses.

H0: _______________
H1: _______________


Write the rejection rule for the null hypothesis.


Calculate the appropriate test statistic.


Write a conclusion. Do we reject or fail to reject the null hypothesis? What do we believe about the fat content of the yogurt?


Find the p-value.

Answer 1

if i had a calc i could do it, but here's how you can do it w/ a TI calc...
H0: yogurt = 1.5 g of fat
H1: Yogurt > 1.5g of fat

perform a 1-Sample t-test

get the p-value

see if the p-value is lower or higher than .1

if it is lower there is enough evidence to reject the null hypothesis of the yogurt containing 1.5g of fat.

if it is higher then there is not enough evidence to reject the null

Answer 2

Assumption: Your data (Life yogurt) come from a normal population with mean 1.5 grams
H0: Yogurt contains 1.5 grams
H1: Yogurt contains more than 1.5 grams
Sample size n=90
mean xbar =1.54
sd = 0.16
mu= population mean of yogurt in grams
z=(xbar-mu)/(s/sqrt(n)) test statistic
Rule: Reject null hypothesis if z > 10% of normal critical value which is 1.28
Find prob( Fat > 1.5 grams)
z = (1.54-1.5)/(0.16/sqrt(90))
sqrt(90)=9.4868
z = (1.54-1.5)/0.0169=0.04/0.0169=2.37 > 1.28
Conclusion: At 10 % sig.level, the prob of obtaining a value as large or larger than 2.37 exceeds the critical value. So, reject the null hypothesis. Product contains more fat than advertised and warrants a recall. Product violated FDA standards.

p-value : prob( z > 2.37) = 0.0089 (From normal table). This is less than 10 % or 0.10 and the null hypothesis is rejected. (This compares p-values rather than the test statstic)

Answer 3

H0: Fat <= 1.5 g
H1: Fat > 1.5g

Using Minitab, I got a p=0.021. Reject null.

Answer 4

Ho: u losing Team Fouls - u Winning Team Fouls