1 NitrateLevel 2 50.1 3 48 4 46 5 45.9 6. 39.9 7 52.1
Inverse Normal Distribution
The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.
Mean, Median, Mode
It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.
Z-Scores
A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.
Answer:
(a) (i)
| 44.52 | confidence interval 95.% lower |
| 48.42 | confidence interval 95.% upper |
| 1.951 | margin of error |
(ii)
| 43.77 | confidence interval 99.% lower |
| 49.18 | confidence interval 99.% upper |
| 2.708 | margin of error |
The 95% confidence interval for the mean amount is between 44.52 and 48.42.
(b) The hypothesis being tested is:
H0: µ = 50
Ha: µ < 50
| 50.0000 | hypothesized value |
| 46.4733 | mean NitrateLevel |
| 3.5233 | std. dev. |
| 0.9097 | std. error |
| 15 | n |
| 14 | df |
| -3.877 | t |
| .0008 | p-value (one-tailed, lower) |
The p-value is 0.0008.
Since the p-value (0.0008) is less than the significance level (0.01), we can reject the null hypothesis.
Therefore, we can conclude that µ < 50.
(c) The hypothesis being tested is:
H0: µ = 50
Ha: µ ≠ 50
| 50.0000 | hypothesized value |
| 46.4733 | mean NitrateLevel |
| 3.5233 | std. dev. |
| 0.9097 | std. error |
| 15 | n |
| 14 | df |
| -3.877 | t |
| .0017 | p-value (two-tailed) |
The p-value is 0.0017.
Since the p-value (0.0017) is less than the significance level (0.01), we can reject the null hypothesis.
Therefore, we can conclude that µ < 50.
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