PR?NTER VERS?ON \'BACX1 A consumer preducts compeny is tormuating shamooe How-17
ID: 3367264 • Letter: P
Question
PR?NTER VERS?ON 'BACX1 A consumer preducts compeny is tormuating shamooe How-175-at Hi: ?> 175 ndimeters, using the resits of na 10 angle. and is inberested in foam height on milimeters). Foam height is approximately i narmally distributed and has a standard deviation of 20 milimaters The company wished to teat the sample data resut in R-190 millmeters, find the value for the test statbatic a Round your answer to twe dedimal places te g. 98.76) the absolute tolerance is/0.0 LINK TO TEXT How unusuat" is the sample valce R- 190 milimeters? Round your anwer to four decimal places (o.g. 98.7654 f the trve meen is realy 1757 That is, what is the probeblity thet you mould observe a sample averege as arge as 130 milsimeters (or lerger), the true mean foam height wes realily 175 is the ebsolube tolerance i +/-0.000 ns.J Al Rights Reserved. A Divisian ofExplanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: u < 175
Alternative hypothesis: u > 175
Note that these hypotheses constitute a one-tailed test.
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)
S.E = 6.32456
DF = n - 1
D.F = 9
t = (x - u) / SE
t = 2.37
where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a t statistic test statistic of 2.37.
Thus the P-value in this analysis is 0.021.
Interpret results. Since the P-value (0.021) is less than the significance level (0.05), we have to reject the null hypothesis.
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