A pond has 150 fish initially, and the population decreases by 8% each day.
1 answer:
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Answer:
We Reject H₀ if t calculated > t tabulated
But in this case,
0.83 is not greater than 2.056
Therefore, we failed to reject H₀
There is no difference between the mean pulse rate of people who do not exercise and the mean pulse rate of people who do exercise.
Step-by-step explanation:
Refer to the attached data.
The Null and Alternate hypothesis is given by
Null hypotheses = H₀: μ₁ = μ₂
Alternate hypotheses = H₁: μ₁ ≠ μ₂
The test statistic is given by
Where is the sample mean of people who do not exercise regularly.
Where is the sample mean of people who do exercise regularly.
Where is the sample standard deviation of people who do not exercise regularly.
Where is the sample standard deviation of people who do exercise regularly.
Where is the sample size of people who do not exercise regularly.
Where is the sample size of people who do exercise regularly.
The given level of significance is
1 - 0.95 = 0.05
The degree of freedom is
df = 16 + 12 - 2 = 26
From the t-table, df = 26 and significance level 0.05,
t = 2.056 (two-tailed)
Conclusion:
We Reject H₀ if t calculated > t tabulated
But in this case,
0.83 is not greater than 2.056
Therefore, We failed to reject H₀
There is no difference between the mean pulse rate of people who do not exercise and the mean pulse rate of people who do exercise.
