91影视

The null hypothesis states that the population mean is equal to the value mentioned in the claim. If the null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

\(\begin{array}{l}{H_0}:\mu = 0\\{H_a}:\mu \ne 0\end{array}\)

The formula for the value of the test statistic is given by, \(t = \frac{{\bar x - {\mu _0}}}{{s/\sqrt n }}\).

The mean is the ration of sum of all values and the total number of values.

\(\begin{aligned}{c}\bar x &= \frac{{85 + 77 + 82 + 68 + 72 + 69}}{6}\\ &= \frac{{453}}{6}\\ &\approx 75.5\end{aligned}\)

The square of the variance is the standard deviation.

\(\begin{aligned}{c}s &= \sqrt {\frac{{{{(85 - 75.5)}^2} + \ldots . + {{(69 - 75.5)}^2}}}{{6 - 1}}} \\ &\approx 7.0071\end{aligned}\)

Let the given be:

\(\begin{array}{l}n = 6\\\alpha = 0.05\end{array}\)

Claim that the average CO concentration is at\(70ppm\).

The value of the test statistic:

\(\begin{aligned}{c}t &= \frac{{\bar x - {\mu _0}}}{{s/\sqrt n }}\\ &= \frac{{75.5 - 70}}{{7.0071/\sqrt 6 }}\\ &\approx 1.923\end{aligned}\)

The P-value is the chance of getting the test statistic's result, or a number that is more severe. The P-value is the number (or interval) in the column header of the T table in the appendix that contains the t-value in the row\(\begin{array}{c}df = n - 1\\ = 6 - 1\\ = 5\end{array}\)for the student.

\(0.10 = 2 \times 0.05 < P < 2 \times 0.10 = 0.20\)

As the P-value is smaller than the significance level, so the null hypothesis is rejected.

\(P > 0.05 \Rightarrow Fail to Reject {H_0}\)

The average CO concentration is at \(70ppm\) is not supported by appropriate evidence.