I\'m not sure what to do because there\'s no information on the sample size, cou
ID: 3352141 • Letter: I
Question
I'm not sure what to do because there's no information on the sample size, could someone help me with this? And explain the answer, thank you!
/pid-985 1077-dt-content-rid-49473091-1/courses/CHEM2250001201830/Exam%201.pdf LetieWatehThis1 MSN.comY Yahoo TwitKing Chet odav Watch ree movie ? Twitter Question 1: The amount of hydrocarbons inside 31 cars was measured on the NJ turnpike and within the Lincoln Tunnel between NJ & NY. The measured concentrations are shown below; Turnpike: Mean concentration: 31.4 mg/L (standard deviation 30.0 mg/L) Tunnel: Mean concentration: 52.9 mg/L (standard deviation 30.0 mg/L) (a) Are the population (standard) deviations the same at the 95% confidence level? (b) If the pooled standard deviation (Spooled) is 29.9, do the results differ at the 95% confidence level?Explanation / Answer
Solution:-
a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 7.62
DF = 62
t = [ (x1 - x2) - d ] / SE
t = - 2.82
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 62 degrees of freedom is more extreme than - 2.82; that is, less than -2.82 or greater than 2.82.
Thus, the P-value = 0.0064
Interpret results. Since the P-value (0.0064) is less than the significance level (0.05), we cannot accept the null hypothesis.
b)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
S.D = 29.9
SE = 7.59
DF = 62
t = [ (x1 - x2) - d ] / SE
t = - 2.83
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 62 degrees of freedom is more extreme than - 2.83; that is, less than -2.83 or greater than 2.83.
Thus, the P-value = 0.0063.
Interpret results. Since the P-value (0.0063) is less than the significance level (0.05), we cannot accept the null hypothesis.