Score: 0.17 of 1 pt 6017 Test Score: 2381% 1 67 017 pts 7.4.33-T Question Help *

Question

Score: 0.17 of 1 pt 6017 Test Score: 2381% 1 67 017 pts 7.4.33-T Question Help * A health joumal conducted a study to see if packaging a healhy food product ike junk food would inflence childrens desire to consume he product A fettous brand of a healthy food prodact-siced appleswas packaged to appeal to children Tha researchers showed the packaging to a sumple of 332 schodl chid en and asked each whother he or she was wiling to eat the produsct wilngness to est was easued at all and 5-very willing The data are summarized as x -351 and s 2.42 Suppose the researchers knew that the mean willingness to eat an actual brand of sliced applies (which is not packaged for children) is p-3 Complete parts a and b below a, conduct a test to dotermne whether the true mean wlngness to eat rg brand of sleed apples packaged br chiireiteeded·U-e010 b multe ror erin on State the null and altemative hypotheses Find the test statissc test statishec 3.84 (Round to two decmal places as needed) Find the p-vale p-vale-0.00o" cRound to three decmal places as needed ) what-th4popriate conclusion at "" 0 107 Repct Ho There is inuficent ovidence to conclude hat the true mean enponse for al kchoole Repct Ho There i s insufficient evidence to conclude that the true mean response for all school children is greater han Quention in complete Tap on the red ndicators to see nconect answers Question is complete Tap on the red indicators to see incoerect answers

Explanation / Answer

Solution:-

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

Null hypothesis: < 3

Alternative hypothesis: > 3

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

Formulate an analysis plan. For this analysis, the significance level is 0.10. 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)

S.E = 0.1328

DF = n - 1 = 332 - 1

D.F = 331

t = (x - ) / SE

t = 3.84

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.

The observed sample mean produced a t statistic test statistic of 3.84. We use the t Distribution Calculator to find P(t > 3.84) = 0.000074

Thus the P-value in this analysis is 0.000074

Interpret results. Since the P-value (0.000074) is less than the significance level (0.10), we have to reject the null hypothesis.

Reject H0, there is sufficiente evidence to conclude that the true mean response for all children is greater than 3.

b) The conclusion is still valid because the sample size is large enough that the central limit theorem applies.

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