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In a study of the impact of eating fruit on birth weight, researchers analyze birth weights (in grams) for babies born to 189 women who gave birth in 2018 at a hospital in California. Suppose in the group, 74 of the women were categorized as “fruit eaters” and 115 as “non-fruit eaters.” The difference in the two sample mean birth weights (fruit eaters minus non-fruit eaters) is 281.7 grams and the 95% confidence interval is (76.5, 486.9)
Which gives the best interpretation of what we can conclude about the impact of eating fruit on birth weight?
Eating fruit is associated with higher birth weights. When fruit eaters are compared to non-fruit eaters, we are 95% confident that the mean weight of babies of fruit eaters is between 76.5 grams to 486.9 grams more than the mean weight of babies of non-fruit eaters.This study is an observational st...
Read MoreCollege Students and Depression: A public health official is studying differences in depression among students at two different universities. They collect a random sample of students independently from each of the two universities and administer a well known depression inventory. A score of 5 or above indicates some depression. A score above 15 indicates that active treatment is necessary.
Sample Statistics
Size(n) | Mean | SD(s) | |
Sample 1 | 50 | 9.2 | .85 |
Sample 2 | 45 | 8.7 | 1.2 |
The official conducts a two-sample t-test to determine whether these data provide significant evidence that students at University 1 are more depressed than students at University 2. The test statistic is t = 2.64 with a P-value 0.005.
Which of the following is an appropriate conclusion?
The samples provide significant evidence that students at University 1 are more depressed than students at University 2.Good job! A P-value this small indicates statistically significant results....
Read MoreAn advertiser is experimenting with a new color scheme and conducts a study to test its effectiveness. In which situation could they use the two-sample t-test for comparing two population means?
They randomly expose consumers to one website when they land on their page: the old one with the original color scheme or the new one with the updated color scheme. Then they measure to see how much people buy.Good Job! Since participants in this study were randomized to two different groups the dat...
Read MoreSuppose the results indicate that the null hypothesis should not be rejected; thus, it is possible that a type II error has been committed.
Given the type of error made in this situation, what could researchers do to reduce the risk of this error?
This is correct. The most straightforward way to avoid type 2 error is to increase the sample size.
None of the above...
Read MoreSuppose the results indicate that the null hypothesis should be rejected; thus, it is possible that a type I error has been committed.
Given the type of error made in this situation, what could researchers do to reduce the risk of this error?
This is correct. The most straightforward way to avoid type 1 error is to choose a more conservative significance level. In this case, .001 is the best option to replace .05 because it is the most conservative.
Change the significance level from .05 to .001This is correct. The most straightforward way to avoid type 1 error is to choose a more conservative significance level. In this case, .001 is the best option to replace .05 because it is the most conservative....
Read MoreA manufacturer of a hair dye markets its color as lasting, on average, 200 washes without fading. A consumer group decides to re-test this claim by assessing the number of times that 30 women can wash their hair without the color fading and finds the average is 200.05 washes, with a standard deviation of 1.5.
The resulting p-value is .82588; thus, the null hypothesis is not rejected. The consumer group concludes that the manufacturer’s claim that its hair color lasts, on average, 200 washes is accurate.
What type of error is possible in this situation?
type IIGood job! The statement describes a situation where we fail to reject a false null hypothesis. This could be a type II error....
Read MoreA manufacturer of a hair dye markets its color as lasting, on average, 200 washes without fading. A consumer group decides to test this claim by assessing the number of times that 30 women can wash their hair without the color fading and finds the average is 203 washes, with a standard deviation of 5.2.
The resulting p-value is 0; thus, the null hypothesis is rejected. The consumer group concludes that the manufacturer’s claim that its hair color lasts, on average, 200 washes is inaccurate.
What type of error is possible in this situation?
type IThe conclusion here is that the null hypothesis is false but the null hypothesis could actually be true. This would be a type I error....
Read MoreBiomedical researchers are testing a cancer treatment to see if it is safe for human use. This can be thought of as a hypothesis test with the following hypotheses.
The following is an example of what type of error?
The sample suggests that the medicine is not safe, but it actually is safe.
type IGood job! The conclusion here is that the null hypothesis is false but the null hypothesis is actually true. This is a type I error....
Read MoreBiomedical researchers are testing a cancer treatment to see if it is safe for human use. This can be thought of as a hypothesis test with the following hypotheses.
The following is an example of what type of error?
The sample suggests that the medicine is safe, but it actually is not safe.
type IIGood job! The statement describes a situation where we fail to reject a false null hypothesis. This is a type II error....
Read MoreThe Wechsler Adult Intelligence Scale (IQ test) is constructed so that Full Scale IQ scores follow a normal distribution, with a mean of 100, and a standard deviation of 15. Dr.Smartyskirt is a University professor and believes that university professors are smarter than the national average but she is worried that too many people see doctors and lawyers as being the smartest professionals. She wants to test to see if doctors and lawyers are smarter than the national average for marketing purposes (she wants to recruit more young people to the PhD track).
A researcher is hired to conduct a study to determine whether doctors and lawyers, on average, have higher Full Scale IQs than the population. A random sample of 100 doctors and lawyers from various practices were given the IQ test and were found to have an average Full Scale IQ of 130.
Which hypothesis test should be used to determine whether the mean Full Scale IQ score of the professors is higher than the national average?
Correct! Since the population mean is being assessed and the population standard deviation is known, the appropriate hypothesis test would be the z-test for the population mean.
What are the null and alternative hypotheses?
What is the value of the test statistic used to determine whether the mean Full Scale IQ score of the doctors and lawyers is higher than the national average?
z-test for the population meanCorrect! Since the population mean is being assessed and the population standard deviation is known, the appropriate hypothesis test would be the z-test for the population mean.H0: μ = 100Ha: μ > 100Correct! The researcher wants to test whether the mean Full Scale IQ...
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