If all city residents are surveyed then a vast majority of them would be opposed

Question

If all city residents are surveyed then a vast majority of them would be opposed to the new tax. If p is the proportion of city residents who are opposed to the new tax, then p = 0.90. The radio program host's survey was a waste of time, and tells nothing useful about the opinion of the city dwellers regarding the new property tax. The callers do not represent either the city dwellers or the program's listeners. None of the above. A random sample of 330 cakes made by a bakery showed that 31 were burnt. After upgrading the baking system, another sample of 278 cakes showed that 11 were burnt. In order to test whether the system change actually improved baking in terms of lower proportion of burnt cakes, the p-value should be (correct up to third decimal value) 0.040: 0.008: 0.094: 0.004: none of these A very common type of avalanche is called the slab avalanche which affects the western part of the United States and Canada. Slab avalanches studied in Canada had an average thickness of mu = 67 cm. The ski patrol at Vail, Colorado, is studying slab avalanches took a random sample and obtained the following thicknesses (in cm): 63, 74, 65, 79, 59, 51, 76, 38, 65, 54, 49, 62, 68, 55, 64, 67 Based on the data, and using 5% level, we can conclude that the mean slab thickness in Vail region is Less than that in Canada Different from that in Canada More than that in Canada (d) All of the above None of the above The General Social Survey asked 1373 men and 993 women in the US whether they agreed that they were generally optimistic about the future. The results are presented as: Male Female Optimistic 1148 815 Pessimistic 225 178 Choose the most appropriate statement Being optimistic is not gender specific Being pessimistic is not gender specific We can run a 2 proportion Z-Test All the above statements are correct None of the above is correct To estimate mu, the mean salary of full professors at American colleges and universities, you obtain the salaries of a random sample of 400 full professors. The sample mean is exist73, 220, and the sample standard deviation is exist4400. A 99% confidence interval for mu is 73, 220 plusminus 28.6 73, 220 plusminus 123.7 73, 220 plusminus 431 73, 220 plusminus 572 73, 220 plusminus 11, 440

Explanation / Answer

6) Here we want to test whether the system change actually improved backing in terms of lower proportion of burnt

cakes. that is proportion of second sample is less than the proportion of first sample

So here alternative hypothesis H1: P1 > P2

This implies H1 : P1 - P2 > 0

from the given information

n1 = number of trials of first sample = 330

x1 = number of success events for first sample = 31

n1 = number of trials of first sample = 278

x1 = number of success events for first sample = 11

using Minitab

Steps : Click on Stat>>>Basic Statistics >>>2 sample proportion

Then select summarized data and put the given data.

then Click on Option

Select alternative greater than

Then click on OK, again click on OK then we get the following output

Test and CI for Two Proportions

Sample X N Sample p
1 31 330 0.093939
2 11 278 0.039568

Difference = p (1) - p (2)
Estimate for difference: 0.0543710
95% lower bound for difference: 0.0216958
Test for difference = 0 (vs > 0): Z = 2.63 P-Value = 0.004

From the above ouput P-Value = 0.004

So correct option of question 6 is (d)

7) For these question we need to do check all the three types of alternative hypothesis.

Here we need to do t test of testing population mea.

Using minitab :

Enter the data in the minitab column

Stat>>> basic statistic>>> 1-sample t...

for two tailed t test the minitab output is

Test of mu = 67 vs not = 67

Variable N Mean StDev SE Mean 95% CI T P
X 16 61.81 10.65 2.66 (56.14, 67.49) -1.95 0.070

Decision rule: 1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.07 > 0.05 so we used 2nd rule.

That is we fail to reject null hypothesis

Conclusion: At 5% level of significance the population mean may not different than 67

For left tailed t test the minitab output is

One-Sample T: X

Test of mu = 67 vs < 67

95% Upper
Variable N Mean StDev SE Mean Bound T P
X 16 61.81 10.65 2.66 66.48 -1.95 0.035

Decision rule: 1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.035 < 0.05 so we used first rule.

That is we reject null hypothesis and accept alternative hypothesis.

Conclusion: At 5% level of significance there are sufficient evidence to say that population mean is less than 67

So correct option of question # 7 is a) less than that in canada

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