Question

A candy manufacturer claims that there is 13.1 grams of candy in a fun sized bag. The 23 random samples are shown in the picture.

a.) Calculate the sample mean, the standard error, and sketch/label the distribution curve.

b.) At the alpha level of 0.05, do these samples provide convincing evidence that fun sized bags are being sold under advertised weight? show the hypothesized test H sub o, alternative hypothesis H sub A, and p value.

c.)What would be a 95% confidence interval foe the sample mean?

d.) Calculate the beta score with the assumption that the sample mean you calculated in part a. is the actual population mean vs. the lower edge calue of the confidence interval (edge of the alpha rejection zone) from the original probability.

e.) what is the power of the test?

f.) what is the probability of a type one error? of a type two error?

g.) What would you recommend about accepting this power to a law firm? what should the firm do?

A candy an Achurch shat there.. is l3.Iaron dfand calculate sapé rean, sfrdard emar, S Sketch lakel this distribuhion (t 2 o /3.75 13.5 12.86 11.46G 12.01 I 2.8 一一 . these Sanples ponde conmineuna evikence hot fon-sited b are bing sold nder the advertised weight shouw He hupohesized test H HA, and pale) 13.76 13.67 3 12.33 21 H.76 12.84 119 lelah daould be a 95% Confidence Internal for 52-11.11 11.98 11.99 1198 1254 12.77 ate β(ben/ with the assumphon that- 11571127 the sample nean you caleuleted in part'a is the .8 the co conidene intenal Ledge of the is rejectian zone) eis th per d le test? Tupe err a) whet woud you profesignaly, eommenl about acceping this puer to a lau 5. /3.75 135 12.86 4 12.01 12.8 13.76 13.67 1292 13.58 38 /283 12.84 19 2.1 76 1199 1198 /254 12.77 197 /2.79 89

Answer 1

Column1
Mean	12.76695652
Standard Error	0.166382619
Median	12.8
Mode	11.99
Standard Deviation	0.797943008
Sample Variance	0.636713043
Kurtosis	0.174706426
Skewness	0.581046861
Range	3.3
Minimum	11.46
Maximum	14.76
Sum	293.64
Count	23
Confidence Level(95.0%)	0.34505643

8 4 14.2 14.9 11.4 12.1 12. 13.5 Histogram (Frequency Diagram)

The provided sample mean is X = 12.76695652 and the sample standard deviation is s = 0.797943008, and the sample size is n 23. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested Ho: 2 13.1 На: < 13.1 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used Based on the information provided, the significance level is a 0.05, and the critical value for a left- tailed test is te1.717 The rejection region for this left-tailed test is R-t t<-1.717 (3)_Test Statistics The t-statistic is computed as follows: X-12.76695652 13.1 s/Vn 0.797943008//23 一一2.002 = -= /

(4) Decision about the null hypothesis Since it is observed that tーー2.002 < tc-一1.717, it is then rejected Using the P-value approach: The p-value is p0.0289, and since p concluded that the null hypothesis is rejected. 0.0289<0.05, it is It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is less than 13.1, at the 0.05 significance level The 95% confidence interval is 12.422 < μ < 13.112.