Question

Use Excel Functions and show which functions you used. You may take screenshots.

For Problems #1 through #9, consider the following information: According to product packaging, the calorie count for a single house brand premium milk chocolate chip pecan cookie baked exclusively for Sale Mart is 200 calories. A consumer health testing lab tech is double checking this claim, and has collected a random sample of 16 individual milk chocolate chip pecan cookies. The corresponding calories were as follows 193 204 196 192 194 186 203 187 198 193 195 198 194 197 190 200 Test the reasonability of the claim by Sale Mart at the 5% level of significance. COMMENTS & HINTS: When calculating sample statistics, round any answers to the nearest whole calories if necessary.] #1: What are the null and alternative hypotheses, stated concisely with proper symbols? #2: What type of hypothesis test is involved: two-tail, one-tail left, or one-tail right? Explain #3: Which type of test statistic is generated (i.e., z or ) and why? Explain. #4: Determine the critical value(s) that designate the rejection region(s). Indicate the critical value(s) on an appropriate probability curve (i.e., z or t) so to represent the test in graph form #5: Determine the value of the test statistic (i.e., z or t). #6: Determine the p-value that corresponds to the test statistic. #7: Concisely state the technical conclusion using appropriate and complete statistical language #8: Interpret the findings in complete thoughts/phrases and within the context of the problem #9: Construct the corresponding confidence interval which affirms the conclusion

Answer 1

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TABLE 6 Critical Values for Student's t Distribution one-tail area 0.250 0.125 0.100 0.075 0.050 0.025 0.010 0.005 two-tail area 0.500 0.250 0.200 0.150 0.100 0.050 0.020 0.010 d.f. 0.500 0.750 0.800 0.850 0.900 0.950 0.980 0.990 .000 2.414 3.078 4.165 6.314 12.706 31.821 63.657 0.816 1.604 1.886 2.282 2.920 4.303 6.965 9.925 0.765 1.423 1.638 1.924 2.353 3.182 4.541 5.841 0.741 1.344 1.533 1.778 2.132 2.776 3.747 4.604 0.727 1.30 1.476 1.699 2.015 2.571 3.365 4.032 0.718 1.273 1.440 1.650 1.943 2.447 3.143 3.707 0.711 1.254 1.415 1.617 1.895 2.365 2.998 3.499 0.706 1.240 1.397 1.592 1.860 2.306 2.896 3.355 0.703 1.230 1.383 1.574 1.833 2.262 2.821 3.250 0.700 1.221 1.372 1.559 1.812 2.228 2.764 3.169 0.697 1.214 1.363 1.548 1.796 2.201 2.718 3.106 0.695 1.209 1.356 1.538 1.782 2.179 2.681 3.055 0.694 1.204 1.350 1.530 1.771 2.160 2.650 3.012 0.692 1.200 .345 1.523 1.761 2.145 2.624 2.977 0.691 1.197 1.34 1.517 1.753 2.131 2.602 2.947 0.690 1.194 .337 1.512 1.746 2.120 2.583 2.921 8 9 10 12 13 14 15

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#5)

The sample mean ( X bar ) = 195

Sample standard deviation ( Sd) = 5.046

n = 16

M = 200

Test statistic :

t= (X bar - M) / (Sd / sqrt(n))
t = (195 -200)/(5.046/sqrt(16)

t = -3.9635

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#6)

The formula of P value is:

df = 16-1 = 15

Two tailed test test

t = -3.9635

use excel function: =t.dist.2t(3.9635,15)

P value = 0.001249

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#7)

Decision rule:

P value < alpha, reject the null hypothesis

We got P value 0.001249 < alpha 0.05 , hence reject the null hypothesis

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#8)

Conclusion: Since the P value < alpha , and the T statistic -3.9635 lies in the rejection region ( < -2.131) , hence one concludes to reject the null hypothesis, which would imply that there is significant evidence to conclude that the mean is not equal to 200 and hence the initial claim made that mean is 200 is not supported by this sample data

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#9)

Formula of confidence interval is:

x bar - t * Sd/ sqrt( n) , X bar + t * Sd / sqrt(n)

195 -2.131 * 5.046/sqrt(16) , 165 +2.131 *5.046/sqrt(16)

(192.3117 , 197.6883)