Homework SourseA Good website design can make Web navigation easier An arti
A- Good website design can make Web navigation easier. An article presents a comparison of item recognition between two designs. A sample of 10 users using a conventional Web design averaged 32.7 items identified, with a standard deviation of 8.35. A sample of 10 users using a new structured Web design averaged 44.1 items identified, with a standard deviation of 9.89. Can you conclude that the mean number of items identified with the new structured design exceeds that of the conventional design by more than 2?
A1- We conclude that the mean number of items identified with the new structured design does not exceed that of the conventional design by more than 2.
A2- We conclude that the mean number of items identified with the new structured design exceeds that of the conventional design by more than 2.
-----------------------------------------------------------------------------------------------------------------------------------------------------------
B- An article reports that in a sample of 357 calves less than 2 days old, 107 had lesions on their bronchial lymph nodes, while in a sample of 466 calves more than 2 days old, 140 had such lesions. Can you conclude that the proportion of calves with lesions differs between the two age groups? Carry four significant digits throughout your calculations.
Since the Pvalue is (--------------------------) , we ( (select) cannot / can ) conclude that the proportion of calves with lesions differs between the two age groups.
------------------------------------------------------------------------------------------------------------------------------------------------------------------
C - Crash testing is a highly expensive procedure to evaluate the ability of an automobile to withstand a serious accident. A simple random sample of 12 small cars were subjected to a head-on collision at 40 miles per hour. Of them 7 were \"totaled,\" meaning that the cost of repairs is greater than the value of the car. Another sample of 14 large cars were subjected to the same test, and 5 of them were totaled. Find a 95% confidence interval for the difference in the proportions of small cars and large cars that are totaled.
parenleftzd , parenrightzd
Carry four significant digits throughout your calculations
------------------------------------------------------------------------------------------------------------------------------------------------------------------
D- Two methods have been developed to determine the nickel content of steel. In a sample of five replications of the first method on a certain kind of steel, the average measurement (in percent) was
X= 3.17 and the standard deviation was sX = 0.042.The average of seven replications of the second method was Y= 3.25 and the standard deviation was sY = 0.20. Assume that it is known that the population variances are nearly equal. Can we conclude that there is a difference in the mean measurements between the two methods? Carry three significant digits throughout your calculations.
D1 - There does not appear to be a difference in the mean measurements between the two methods.
D2 - There does appear to be a difference in the mean measurements between the two methods.
-------------------------------------------------------------------------------------------------------------------------------------------------------------
E- A random sample of 320 electronic components manufactured by a certain process are tested, and 26 are found to be defective. Let p represent the proportion of components manufactured by this process that are defective. The process engineer claims that p 0.05. Does the sample provide enough evidence to reject the claim?
E1 - Yes, the sample provides enough evidence to reject the claim.
E2 -No, the sample does not provide enough evidence to reject the claim.
------------------------------------------------------------------------------------------------------------------------------------------------------------------
F- Two weights, each labeled as weighing 100 g, are each weighed several times on the same scale. The results, in units of g above 100 g, are as follows:
First weight: 53 89 89 63 39 66
Second weight: 23 39 28 2 49
Since the same scale was used for both weights, and since both weights are similar, it is reasonable to assume that the variance of the weighing does not depend on the object being weighed. Can you conclude that the weights differ?
Since the Pvalue is (--------------------------) , ( we (select) cannot\\ can ) conclude that the weights differ.
Solution
A.
Formulating the null and alternative hypotheses,
Ho: u1 - u2 <= 2
Ha: u1 - u2 > 2
At level of significance = 0.05
As we can see, this is a right tailed test.
Calculating the means of each group,
X1 = 44.1
X2 = 32.7
Calculating the standard deviations of each group,
s1 = 9.89
s2 = 8.35
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 10
n2 = sample size of group 2 = 10
Thus, df = n1 + n2 - 2 = 18
Also, sD = 4.09309907
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = 2.29654837
where uD = hypothesized difference = 2
Now, the critical value for t is
tcrit = 1.734063607
Thus, comparing |t| > 1.734,, we decide to WE REJECT THE NULL HYPOTHESIS.
Also, using p values,
p = 0.016931912
Comparing this to the significance level, WE REJECT THE NULL HYPOTHESIS.
Therefore:
A2- We conclude that the mean number of items identified with the new structured design exceeds that of the conventional design by more than 2. [ANSWER]
*******************************************
Hi! Please submit the next part as a separate question. That way we can continue helping you! Please indicate which parts are not yet solved when you submit. Thanks!
