needing help with 2, 4, 6, and 8 and 12 alternative hypothesis 406 (beta) 411 ch
ID: 3295095 • Letter: N
Question
needing help with 2, 4, 6, and 8 and 12
alternative hypothesis 406 (beta) 411 chi-square test 455 critical or rejection region 41:2 hypothesis testing 40% left-tailed test 412 level of significance 411 righ-tailed test 412 noncritical or nonrejection statistical hypothesis 4pe r ses 434 wo-tailed tex research hypothesis 400 56 region 412 null hypothesis 406 eype 1l emox 4 z tess 419 statistical test 409 Important Formulas Formula for the z test for means: Formula for the z test for proportions /V if nExplanation / Answer
Solution:-
2)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: > 25.3
Alternative hypothesis: < 25.3
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 0.01. 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)
S.E = 0.639
DF = n - 1 = 100 - 1
D.F = 99
t = (x - ) / SE
t = - 2.19
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.
The observed sample mean produced a t statistic test statistic of - 2.19. We use the t Distribution Calculator to find P(t < - 2.19) = 0.0154
Thus the P-value in this analysis is 0.0154
Interpret results. Since the P-value (0.0154) is greater than the significance level (0.01), we cannot reject the null hypothesis.
From the above test we do not have sufficient evidence in the favor of the claim that mean computing time is less than this particular company.
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