Question

Many consumers pay careful attention to stated nutritional contents on packaged foods when making purchases, so it is important that the information on packages be accurate. The distribution of calorie content has been shown to be approximately normally distributed. A random sample of 12 frozen dinners of a certain type was selected and the calorie content of each one was determined to be 255, 244, 239, 242, 265, 245, 259, 248, 225, 226, 251, and 232. (a) Determine the sample mean and sample standard deviation for these 12 randomly selected frozen dinners. (b) The stated mean calorie content on the box is 240. Construct an appropriate test to see if the actual mean content differs from the stated value at an α = 0.05 significance level. Do not use the T-test feature of your calculator.

Answer 1

Ans(a)

X={255, 244, 239, 242, 265, 245, 259, 248, 225, 226, 251,232}

The formula for mean is:  

this is the average(mean) calorie content in the 12 frozen dinners.

The standard deviation:   

this the standard deviation for the sample.

Ans(b)

SInce our sample size is small we can use the t-statistics(n<30) .

we use t-statistics when we do not know the population standard deviation , but we assume, that we know the population mean.

Before proceeding we have to define the Hypothesis.

, (this is the null hypothesis that actual mean calorie content is 240.)

, (this is our alternative hypothesis, that mean calorie content is not 240.)

Now we now have to calculate the t-statistics:

sample mean

hypothesised mean, or actual mean which we stated in the null hypothesis.

standard error of the sample statistics.

Our observed sample mean of 244.25 calorie content is 1.1807 standard error below 240 (which is our hypothesized true mean value of actual calorie content).

Now at significance level of , we have to find the critical value of t ()

degree of freedom(df)=n-1=12-1=11

, and the p-value= 0.2626,

i.e.,

Since out is smaller than , so we cannot reject the null hypothesis.

As we know if,

, then we reject the Null hypothesis, and say that we have evidence to believe, that or alternative hypothesis is likely true.

, then we FAIL TO REJECT (null hypothesis) and say that we do not have evidence to believe that is likely true.

So we have two conclusion:

(1)

(2)

Conclusion:

So as we cannot reject our null hypothesis, and conclude that we have evidence to support our ( that is the actual mean calorie content is equal to the stated calorie content.)