What is(are) the critical Value(s) Calculate the standardized test statistic Use
ID: 3264115 • Letter: W
Question
What is(are) the critical Value(s)
Calculate the standardized test statistic
Use technology to calculate the p value
Decide wether to reject or fail to reject the null hypotheses
A used car dealer says that the mean price of a three-year-old sports utility vehicle is $20,000. You suspect this claim is incorrect and find that a random sample of 22 similar vehicles has a mean price of $20,636 and a .01 ? Complete parts a trough e belo . Assume he population is normally distri uted. standard deviation of $1982. Is there enough evidence to reject the claim at (a) Write the claim mathematically and identity Ho and Ha- Which of the following correctly states Ho and Ha? Ho: = S20.000 Ha: $20,000 Ha : $20,000 O B. Ho: =$20,000 Ha : > $20,000 0 E. Ho: #820,000 Hai-$20,000 (y c, Ho : -$20,000 Ha : $20,000 F Ho: H2 $20,000 Ha :Explanation / Answer
Solution:-
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 20,000
Alternative hypothesis: 20,000
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n) = 1982 / sqrt(22) = 422.56
DF = n - 1 = 22 - 1 = 21
t = (x - ) / SE = (20636 - 20000)/422.56 = 1.505
Critical value of t0 = + 2.83135956 or (- 2.831 , + 2.831)
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the 0sample size.
Since we have a two-tailed test, the P-value is the probability that the t statistic having 21 degrees of freedom is less than -1.505 or greater than 1.505.
We use the t Distribution Calculator to find P(t < -1.505), and P(t > 1.505)
The P-Value is 0.147217.
The result is not significant at p < 0.01.
Interpret results. Since the P-value (0.147217) is greater than the significance level (0.01), we cannot reject the null hypothesis.
Conclusion. Fail to reject the null hypothesis.
Second question:-
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: <= 30
Alternative hypothesis: > 30
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n) = 4.6 / sqrt(25) = 0.92
DF = n - 1 = 25 - 1 = 24
t = (x - ) / SE = (32 - 30)/0.92 = 2.1739
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Here is the logic of the analysis: Given the alternative hypothesis ( > 30), we want to know whether the observed sample mean is large enough to cause us to reject the null hypothesis.
The observed sample mean produced a t statistic test statistic of 2.1739. We use the t Distribution Calculator to find P(t > 2.1739)
The P-Value is 0.019901.
The result is significant at p < 0.10.
Interpret results. Since the P-value (0.019901) is less than the significance level (0.10), we cannot accept the null hypothesis.
Conclusion. Reject the null hypothesis.
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