A special diet is intended to reduce systolic blood pressure among patients diagnosed with stage 2 hypertension. If the diet is effective, the target is to have the average systolic blood pressure of this group be below 150. After six months on the diet, an SRS of 28 patients had an average systolic blood pressure of = 143, with standard deviation s = 21. Is this sufficient evidence that the diet is effective in meeting the target? Assume that the distribution of the systolic blood pressure for patients in this group is approximately Normal with mean μ. Based on the data, the value of the one-sample t statistic is:
–1.76.
0.33.
–0.33.
-1.76
To determine the value of the one-sample t statistic, we need to calculate the test statistic using the given information. The one-sample t statistic is calculated using the formula:
t = (x̄ - μ) / (s / √n)
where:
- x̄ is the sample mean (143 in this case),
- μ is the population mean (target systolic blood pressure of 150 in this case),
- s is the sample standard deviation (21 in this case),
- n is the sample size (28 in this case).
Let's substitute the given values into the formula:
t = (143 - 150) / (21 / √28)
t = -7 / (21 / 5.29)
t = -7 / 3.97
Calculating this expression, we find that the value equals approximately -1.76.
Therefore, the value of the one-sample t statistic is -1.76.
To determine the value of the one-sample t statistic, we need to calculate the t-value using the given information.
The formula for calculating the t statistic is:
t = (x̄ - μ) / (s / √n)
Where:
x̄ = sample mean = 143
μ = population mean (target) = 150
s = standard deviation = 21
n = sample size = 28
Plugging in the values:
t = (143 - 150) / (21 / √28)
Calculating the value:
t = -7 / (21 / 5.29)
t ≈ -7 / 3.97
t ≈ -1.76
Therefore, the value of the one-sample t statistic is approximately -1.76.