The mean sleep time of babies is 14hrs,a sample of 64 babies showed a mean of 13hrs 30min with a standard deviation of 3hrs at 1% level of significance. Test the assertion that the mean sleep of the babies is less than 14hrs a day.

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of the Z score. Is it more or less than .01?

Learn Hypothesis Step By Step

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To test the assertion that the mean sleep of the babies is less than 14 hours a day, we can use a one-sample t-test. Here are the steps to perform the test:

Step 1: State the null and alternative hypotheses:
- Null hypothesis (H0): The mean sleep of the babies is equal to 14 hours a day.
- Alternative hypothesis (Ha): The mean sleep of the babies is less than 14 hours a day.

Step 2: Define the level of significance (α):
In this case, the level of significance is given as 1%, which means α = 0.01.

Step 3: Calculate the test statistic:
The test statistic for a one-sample t-test is calculated using the formula:
t = (x̄ - μ) / (s / √n)
Where:
- x̄ = sample mean
- μ = population mean (in this case, 14 hours)
- s = sample standard deviation
- n = sample size

In this case, the sample mean (x̄) is 13.5 hours, the population mean (μ) is 14 hours, the sample standard deviation (s) is 3 hours, and the sample size (n) is 64.

Substituting the values into the formula, we get:
t = (13.5 - 14) / (3 / √64)

Step 4: Determine the critical value:
Since we are conducting a one-tailed test (less than), we need to find the critical t-value that corresponds to a 1% significance level and n-1 degrees of freedom. In this case, the degrees of freedom is 63 (n-1).

Using a t-table or a statistical software, the critical t-value is approximately -2.660.

Step 5: Make a decision:
If the absolute value of the calculated test statistic is greater than the critical t-value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

In this case, the calculated t-value needs to be less than -2.660 in order to reject the null hypothesis.

Step 6: Calculate the p-value:
The p-value represents the probability of observing a t-value as extreme as the one calculated, assuming the null hypothesis is true. To find the p-value, we can use a t-distribution table or a statistical software.

Step 7: Make a decision:
If the p-value is less than the significance level (α), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

That's how you can test the assertion that the mean sleep of the babies is less than 14 hours a day using a one-sample t-test.