Observed significance level

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

1. The data below represents a random sample of 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 interval is [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 mean and standard deviation for 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 error of 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 sampling distribution.

Refer to the Standard normal distribution tables 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.