An oceanographer wants to check whether the depth of the ocean in a certain region is 57.4 fathoms,as has previously been recorded. What can he conclude at the 0.05 level of significance,if readings taken at 40 random locations in the given region yield a mean of 59.1 fathoms with a standard deviation 5.2 fathoms
To determine what the oceanographer can conclude, we will perform a one-sample t-test.
Null hypothesis: The true population mean depth of the ocean in the given region is 57.4 fathoms.
Alternative hypothesis: The true population mean depth of the ocean in the given region is not 57.4 fathoms.
We will use the following information:
Sample mean (x̄) = 59.1 fathoms
Sample standard deviation (s) = 5.2 fathoms
Sample size (n) = 40
Population mean under the null hypothesis (μ) = 57.4 fathoms
Significance level (α) = 0.05 (or 5%)
Since the sample size (n) is greater than 30, we can assume that the sampling distribution of the mean is approximately normally distributed.
First, we need to calculate the t-statistic using the formula:
t = (x̄ - μ) / (s / √n)
t = (59.1 - 57.4) / (5.2 / √40)
t = 1.7 / (5.2 / 6.3246)
t = 1.7 / 0.8234
t ≈ 2.064
Next, we need to determine the critical value for a two-tailed test at a 0.05 significance level. Since the sample size is greater than 30, we can use a z-distribution to approximate the critical value.
Using a z-table or calculator, the critical z-value for a 0.025 significance level (divided equally between the two tails) is approximately 1.96.
Since our test statistic (t = 2.064) is greater than the critical z-value (1.96), we reject the null hypothesis.
Conclusion: The oceanographer can conclude that the depth of the ocean in the given region is significantly different from the previously recorded depth of 57.4 fathoms at the 0.05 level of significance.