Many consumers pay careful attention to stated nutritional contents on packaged

Question

Many consumers pay careful attention to stated nutritional contents on packaged foods when making purchases. It is therefore important that the information on packages be accurate. A random sample of n = 12 frozen dinners of a certain type was selected from production during a particular period, and the calorie content of each one was determined. (This determination entails destroying the product, so a census would certainly not be desirable!) Here are the resulting observations, along with a boxplot and normal probability plot.

255 244 239 242 265 245 259 248 225 226 251 233

Carry out a formal test of the hypotheses suggested in part (b). (Use Table 4 in Appendix A. Use = 0.05. Round your test statistic to two decimal places and your P-value to three decimal places.)

t =

df =

P =

Explanation / Answer

Here in the questions some informations are missing, like the population mean should be mentioned, however here is all other information that will help you to get the hypothesis test done.

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: = population mean
Alternative hypothesis: population mean

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n) = 12.38278 / sqrt(12) = 3.5746
DF = n - 1 = 12 - 1 = 11
t = (x - ) / SE = (244.33 - population mean)/3.5746

[Say the given population mean is 246]

So, t = (244.33 - 246)/3.5746 = -0.4672

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 11 degrees of freedom is less than -0.4672 or greater than 0.4672.

We use the t Distribution Calculator.

The P-Value is 0.649614.
The result is not significant at p < 0.05.

Interpret results. Since the P-value (0.649614) is greater than the significance level (0.05), we cannot reject the null hypothesis.

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