Answer:
1) μ = 8000
2) H0: μ ≥8000
3) Ha: μ < 8000
4) α= 0.10
5) If the sample size is at least 15 a student's t test can be used.And we assume that the variances of the two independent groups are almost equal.
6) t= x`- μ/ s.d/ √n = -0.816
7) The P- value = 0.214083.
8) The critical region is [-∞, -1.345]
9) Fail to Reject the Null Hypothesis
Step-by-step explanation:
1) The population characteristic is the population mean =μ = 8000
2) The claim is that the addition of nitrates results in a decrease in the mean rate of uptake i.e μ < 8000
The null hypothesis is that μ ≥ 8000
2) Null hypothesis: H0: μ ≥8000
3) Alternative hypothesis: Ha: μ < 8000
4) The significance level alpha = α= 0.10
5) If the sample size is at least 15 a student's t test can be used.And we assume that the variances of the two independent groups are almost equal.
Calculating mean and standard deviation using a calculator
Sample Size, n: 15
Mean= ∑ x/n = x`=: 7788.8
∑ x/n= 116832/15
St Dev, s.d(n-1) : 1002.4307
6) Test Statistic: student's t Test with d.f = n-1 = 14
t= x`- μ/ s.d/ √n
t= 7788.8-8000/1002.4307/√15
t= -0.816
7) The P- value for the observed t value is 0.214083.
The result is not significant at p < 0.10. H0 is accepted.
8) The Critical value for one sided t test at α =0.1 with 14 d.f is less than t= -1.3406
Hence the critical region is [-∞, -1.345]
Since the calculated t -0.816 value does not fall in the critical region [-∞, -1.345] there is not sufficient evidence to reject H0
9) Fail to Reject the Null Hypothesis
Sample does not provide enough evidence to support the null hypothesis and conclude that the data suggest that the addition of nitrates do not result in a decrease in the mean rate of uptake. The claim is rejected.
