Brief Introduction:
AJ Davis is a department store chain, which has many credit customers and want to find out more information about these customers. AJ Davis has complied a sample of 50 credit customers with data selected in the following variables: Location, Income (in $1,000’s), Size (Number of people living in the household), Years (number of years the customer has lived in the current location), and Credit Balance (customers current credit card balance on the store’s credit car, in $).
The manager at AJ Davis has speculated the following: a. The average (mean) annual income was less than $50,000. b. The true population proportion of customers who live in an urban area exceeds 40% c. The average (mean) number of
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* My calculated test statistic of -2.07 falls in the rejection region of z < -0.1645, therefore, I would reject the null hypothesis and say there is sufficient evidence to indicate u<50 or $50,000.
b. The true population proportion of customers who live in an urban area exceeds 40%
* 22 of the 50 surveyed live in the Urban area, which is 44% or 0.44, this is the point estimate for p.
* Therefore my hypothesis would be * Ho: = 0.40 vs. Ha: p>0.40
* In order to conduct the large sample z-test, we first need to verify that the sample size is large enough. * nPo= 50(0.40) = 20 and 50 (1-0.44) = 30, both are larger than 15, so we can conclude that sample size is large enough to apply the large sample z test.
* Z= (0.44 – 0.400)/ 0.69282= 0.58 where s phat= sqrt (((0.40) (0.60))/50= 0.069282
* This is a one tailed (upper or right since HA has “>”). Our rejection regions would be z > 1.645.
* 0.58 is not greater than 1.645 (and is not in the rejection regions) so we would not Reject the Ho.
* The p-value= 0.282. The p-value is another complementary and equally valid way we can evaluate the null and alternative hypotheses is by looking at the p-value and compare the p-value to alpha. If the p-value is less than alpha, reject the null hypothesis and accept the alternative
The critical value for for this two-tailed test is. The rejection region is given by
From the above output, we can see that the p-value is 0.000186, which is smaller than 0.05 (if we select a 0.05 significance level).
(b) The value of Z with an area of 5% in the right tail, but not the sample mean.
The null hypothesis is rejected since the p-value is below the significance level of 0.05.
a. What is a Z score for a car with a price of $ 33,000?
Assume 20% of all email is spam. A large Internet provider plans on conducting a survey of 900 emails to see what percentage are spam.
Results from previous studies showed 79% of all high school seniors from a certain city plan to attend college after graduation. A random sample of 200 high school seniors from this city reveals that 162 plan to attend college. Does this indicate that the percentage has increased from that of previous studies? Test at the 5% level of significance.
It can be concluded that some variables such as Income are strongly related to the credit balance of AJ DAVIS department store customers. Several other variables appear to be unrelated.
1. A researcher is interested in whether students who attend private high schools have higher average SAT Scores than students in the general population. A random sample of 90 students at a private high school is tested and and a mean SAT score of 1030 is obtained. The average score for public high school student is 1000 (σ= 200).
• Alternative Hypothesis Ha: µ ≥ 50,000, this means the test is a one tailed z test.
AJ Davis is a department store chain, which has many credit customers and wants to find out more information about these customers. The total sample set of 50 credit customers is selected with data collected.
g. Determine the observed p-value for the hypothesis test and interpret this value. What does this mean?
3.|In the figure below, if the -test value is 1.43, the null hypothesis should not be rejected. |
5) From calculations, computed z value is more than -1.65 and falls within Ho not rejected region. Ho is not rejected at α = 0.05 & α = 0.01 significance levels.
Here P value is (0.695/2)= 0.3475 which is grater than significance level 0.01. If α> P value reject, do not reject otherwise, here 0.3475 ( P value) is grater than significance level, so do not reject. That means we can conclude that there is not enough evidence to claim that, the mean selling price in the Gulshan area is more than