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:
<u><em>1) The population characteristic</em></u> 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
<u><em>2) Null hypothesis:</em></u> H0: μ ≥8000
<u><em>3) Alternative hypothesis:</em></u> Ha: μ < 8000
<u><em>4) The significance level</em></u> alpha = α= 0.10
<u><em>5) If the sample size</em></u> 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
<u><em>6) Test Statistic</em></u>: 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
<u><em>7) The P- value</em></u> for the observed t value is 0.214083.
The result is not significant at p < 0.10. H0 is accepted.
<u><em>8) The Critical value</em></u> 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
<u><em>9) Fail to Reject the Null Hypothesis</em></u>
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.
