Question

test a claim that the mean amount of carbon monoxide in the air in u.s. cities is less than 2.31 parts per million. it was found that the mean amount of carbon monoxide in the air for the random sample of 65 cities is 2.30 parts per million and the standard deviation is 2.11 parts per million. at \\(\alpha = 0.01\\), can the claim be supported? complete parts (a) through (e) below. assume the population is normally distributed.

(c) find the standardized test statistic, \\(t\\)
the standardized test statistic is \\(t = 0.27\\)
(round to two decimal places as needed.)

(d) decide whether to reject or fail to reject the null hypothesis
fail to reject \\(h_0\\) because the standardized test statistic is not in the rejection region.

(e) interpret the decision in the context of the original claim.
there is not enough evidence at the 0.01% level of significance to support the claim that the mean amount of carbon monoxide in the air in u.s. cities is less than or equal 2.31 parts per million
(type integers or decimals. do not round.)

Explanation:

Step1: Identify the test type

This is a one - sample t - test (since population standard deviation is unknown, we use sample standard deviation, and population is normally distributed). The formula for the t - statistic is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step2: Plug in the values

Given \(\bar{x} = 2.30\), \(\mu=2.31\), \(s = 2.11\), \(n = 65\). First, calculate the denominator \(s/\sqrt{n}=\frac{2.11}{\sqrt{65}}\approx\frac{2.11}{8.0623}\approx0.2617\). Then, the numerator \(\bar{x}-\mu=2.30 - 2.31=- 0.01\). Then \(t=\frac{- 0.01}{0.2617}\approx - 0.04\) (Wait, the given value in the problem is 0.27, maybe there is a miscalculation in the problem's given or in my understanding. Wait, maybe the formula was misapplied. Wait, if we consider the claim is \(\mu<2.31\), null hypothesis \(H_0:\mu = 2.31\), alternative \(H_a:\mu<2.31\). Then \(\bar{x}=2.30\), \(\mu = 2.31\), \(s = 2.11\), \(n = 65\). So \(t=\frac{2.30 - 2.31}{2.11/\sqrt{65}}=\frac{- 0.01}{2.11/8.0623}=\frac{-0.01}{0.2617}\approx - 0.04\), but the problem says 0.27. Maybe there was a typo in the sample mean or standard deviation. But let's proceed with the decision.

For part (d), we need to find the critical value for a left - tailed test with \(\alpha = 0.01\) and \(df=n - 1=64\). The critical value \(t_{\alpha,df}=t_{0.01,64}\approx - 2.389\) (using t - table or calculator). The test statistic (given as 0.27, but if it's - 0.04 or 0.27, since it's greater than - 2.389 (for left - tailed), we fail to reject \(H_0\).

For part (e), since we fail to reject \(H_0\), there is not enough evidence at the 0.01 level of significance to support the claim that the mean amount of carbon monoxide in the air in U.S. cities is less than 2.31 parts per million.

Answer:

For part (c), if we recalculate correctly, \(t=\frac{2.30 - 2.31}{2.11/\sqrt{65}}\approx - 0.04\), but the problem's given is 0.27 (possible error). For part (d), Fail to reject \(H_0\) because the standardized test statistic is not in the rejection region. For part (e), There is not enough evidence at the 0.01 level of significance to support the claim that the mean amount of carbon monoxide in the air in U.S. cities is less than 2.31 parts per million.