For each of the below questions (1, 3, 5), determine the observed significance level. How would I do this?

1. The databelow represents a random sampleof weights (in grams) from a batch of a particular product. The nominal value is 87.5 grams. Prove that this sample is consistent with the nominal value, to a 90% confidence level.

Ans:

Mean=86.4

Standard deviation = 5.333

The degree of freedom is 20-1=19.

The critical t value at 0.1 significance level is 1.73

Margin of error=critical value*standard deviation/sqrt(n)=1.73*5.333/sqrt(20)=2.1

Upper limit: 86.4+2.1=88.5

Lower limit: 86.4-2.1=84.3

Therefore, the 90% confidence intervalis [84.3, 88.5]

Since the value 87.5 is within this interval, this sample is consistent with the nominal value, to a 90% confidence level.

Data: (85.81274288, 88.7236203, 83.33202174, 87.39825986, 80.80214318, 79.35480832, 80.6169453, 72.96436719, 90.04840768, 91.57450896, 88.99970927, 87.72667891, 86.87734579, 89.72700899, 94.49946214, 95.43705338, 83.83021658, 88.22971068, 83.5693944, 88.4552547)

3. A manufacturer claims that a particular kind of solar shade can reduce the ambient temperature by 2 degrees centigrade over traditional materials. Over random times over several days, temperatures were taken at the same time under a traditional awning and an adjacent awning using this brand of solar shade. In the Data file below, column A is the traditional and column B is the new shade. Prove whether or not the manufacturer’s claims are true to 95% confidence level.

Ans:

First figure out the difference for each, then figure out the meanand standard deviationfor the difference.

Mean = 2.10168

Standard deviation = 0.52158

Degrees of freedom: 20-1=19

The critical value for 95% confidence interval is 2.09

Margin of error = 2.09*0.52158/sqrt (20) = 0.24375

Lower limit: 2.10168-0.24375=1.85793

Upper limit: 2.10168+0.24375=2.34543.

Since the interval [1.85793, 2.34543] contains the value of 2, we could not conclude that the manufacturer’s claims are true to 95% confidence level.

A B

24.24313 21.87521

28.72897 28.0444

21.48468 19.51498

26.08379 24.38939

25.78293 23.63512

24.10804 21.26163

30.84396 28.50113

18.90814 16.52972

21.3339 19.36128

24.68048 22.21857

29.16487 27.20792

19.73482 17.45729

15.69726 12.61739

21.36029 18.98967

31.69022 30.10404

20.35358 18.84266

24.07602 22.27601

26.95173 24.68808

14.39104 11.87781

24.77026 22.96222

5. A random sample of 135 dentists indicated that 84 recommended sugarless gum for their patients that chew gum. What is a 99% confidence interval for the true proportion of dentists in the population that would make such a recommendation?

ANS:

=Proportion of doctors who recommended sugarless gum=84/135=0.62222

1-0.62222=0.37778

n=sample size=135

Since we are not given population parameters, we will approximate them with sample statistics.

p= =0.62222

q= 0.37778

Estimated standard errorof proportion=S.E = 0.041728

Since sample size is large, we will use z statistics. A 99% confidence level will include 49.5% of the area on either side of the mean in the samplingdistribution.

Refer to the Standard normal distributiontables and look for area=0.475

We get Z=2.58.

Lower limit of confidence interval= 0.62222-2.58*0.041728=0.51456

Upper limit of confidence interval= 0.62222+2.58*0.041728=0.72988

We can say with 99% confidence that proportion of true proportion of the doctors who recommended sugarless gum lies between 0.51456 and 0.712988.

For Question 1, as the level of confidence is 90%, the level of …

The expert examines the observed significance level.