Question

19.Before 1990, there was a higher incidence of birth defects among pregnant mothers. Scientists hypothesized that was because the pregnant women were not getting sufficient amounts of folic acid in their diet. In 1985, the average folic acid intake of an American pregnant woman was 150 micrograms/day. After folic acid fortification of cereals and flour in the late 1990s, it was hypothesized that women were now taking in substantially more folic acid daily. A dietary survey of a sample of 49 pregnant women showed that their daily intakes in the year 2001 were 425 micrograms/day, with a standard deviation of 175 micrograms. Did these pregnant women take in statistically MORE folic acid in their diet in 2001 compared to 1985? Run the appropriate test at a 5% level of significance. Choose the correct response for Step 5: Conclusion of this hypothesis test.

Group of answer choices

Do NOT reject the null because (150-425)/(175/7) is LESS than 1.96

Do NOT reject the null hypothesis because (150-425)/(175/7) is LESS than 1.645.

Reject the null hypothesis because (425-150)/(175/7) > 1.645, p<0.001.

Reject the null hypothesis because (425-150)/(175/7) > 1.96, p<0.001

18.Before 1990, there was a higher incidence of birth defects among pregnant mothers. Scientists hypothesized that was because the pregnant women were not getting sufficient amounts of folic acid in their diet. In 1985, the average folic acid intake of an American pregnant woman was 150 micrograms/day. After folic acid fortification of cereals and flour in the late 1990s, it was hypothesized that women were now taking in substantially more folic acid daily. A dietary survey of a sample of 49 pregnant women showed that their daily intakes in the year 2001 were 425 micrograms/day, with a standard deviation of 175 micrograms. Did these pregnant women take in statistically LESS folic acid in their diet in 2001 compared to 1985? Run the appropriate test at a 5% level of significance. Choose the correct response for Step 5: Conclusion of this hypothesis test.

Group of answer choices

Reject the null hypothesis because (150-425)/(175/7) is < -1.645.

Do NOT reject the null hypothesis because (425-150)/(175/7) is NOT <-1.645.

Do NOT reject the null hypothesis because (150-425)/(175/7) > 1.645.

Reject the null hypothesis because (425-150)/(175/7) <- 1.645, p<0.001.

On average, a person who lives in Switzerland consumes the most chocolate each year compared to people in other countries, at 22.36 pounds per person. A sample of 100 Austrians shows they are not far behind, consuming an average of 20.13 pounds per person with a standard deviation of 3.6 pounds. Is the average amount of chocolate consumed by the Austrians significantly DIFFERENT from the amount consumed by those in Switzerland? Run the appropriate test at a 5% level of significance. Choose the correct response for Step 5: Conclusion of this hypothesis test.

Group of answer choices

Reject the null because (20.13-22.36)/(3.6/10) < -1.96, p<0.001.

Reject the null because (22.36-20.13)/(3.6/10) >1.645, p<0.001.

Reject the null because (22.36-20.13)/(3.6/10) >1.96, p<0.001.

Reject the null because (20.13-22.36)/(3.6/10) < -1.645, p<0.001.

17.On average, a person who lives in Switzerland consumes the most chocolate each year compared to people in other countries, at 22.36 pounds per person. A sample of 100 Austrians shows they are not far behind, consuming an average of 20.13 pounds per person with a standard deviation of 3.6 pounds. Is the average amount of chocolate consumed by the Austrians significantly DIFFERENT from the amount consumed by those in Switzerland? Run the appropriate test at a 5% level of significance. Choose the correct response for Step 5: Conclusion of this hypothesis test.

Group of answer choices

Reject the null because (20.13-22.36)/(3.6/10) < -1.96, p<0.001.

Reject the null because (22.36-20.13)/(3.6/10) >1.645, p<0.001.

Reject the null because (22.36-20.13)/(3.6/10) >1.96, p<0.001.

Reject the null because (20.13-22.36)/(3.6/10) < -1.645, p<0.001.

Answer 1

19.

This is a question of hypothesis testing for the population. We are comparing against the null mean of 1985 = 150. Since the population SD is given and n > 30, we can use z-test.

The sample statistic is for the year 2001. We are testing for the right sided one tailed test.

= 425 = 175 n = 49   = 150

= 150

> 150    ​​​​

Test Stat =

Critical value = ..................................using normal distribution tables

Null	150
Mean	425
SD	175
n	49
Test Stat	11
level of significance	0.05
Critical value	1.6449
p-value	0.0000
Decision	Reject null hypothesis

Reject the null hypothesis because (425-150)/(175/7) > 1.96, p<0.001

18.

This is a question of hypothesis testing for the population. We are comparing against the null mean of 1985 = 150. Since the population SD is given and n > 30, we can use z-test.

The sample statistic is for the year 2001. We are testing for the left sided one tailed test.

= 425 = 175 n = 49   = 150

= 150

< 150    ​​​​

Test Stat =

Critical value = ..................................using normal distribution tables

Null	150
Mean	425
SD	175
n	49
Test Stat	11
level of significance	0.05
Critical value	-1.6449
p-value	0.0000
Decision	Reject null hypothesis

Reject the null hypothesis because (425-150)/(175/7) <- 1.645, p<0.001.

This will give the same answer as above because the statistics are same. Only the C.V. changes the sign due to the different side.

17.

This is a question of hypothesis testing for the population. We are comparing against the null mean of Switzerland = 150. Since the population SD is given and n > 30, we can use z-test.

The sample statistic is for Austria. We are testing for the two tailed test because we are checking for difference.

= 20.13  = 3.6 n = 100   = 22.36

= 22.36

22.36 ​​​​

Test Stat =

Critical value = ..................................using normal distribution tables

Null	22.36
Mean	20.13
SD	3.6
n	100
Test Stat	-6.1944
level of significance	0.05
Critical value	-1.9600
p-value	0.0000
Decision	Reject null hypothesis

We see if |Test Stat | > C.V. for rejecting the null hypothesis.

Reject the null because (20.13-22.36)/(3.6/10) < -1.96, p<0.001.