Questions: Suppose that at Northside High, the number of hours per week that seniors spend on homework is approximately Normally distributed with a mean of 18.6 hours and a standard deviation of 6 hours. For simple random samples of size n=36 from this population, you determine that the sampling distribution of the sampling distribution of the sample mean is approximately normal with a mean of 18.6 hours and a standard deviation of 6/sqrt(36)=1 hour. If you discover that Northside High has only 200 seniors, which of the following would no longer be correct? - the mean and the standard deviation - the standard deviation and the shape of the distribution - All three would be incorrect. - only the standard deviation - only the shape of the distribution

Suppose that at Northside High, the number of hours per week that seniors spend on homework is approximately Normally distributed with a mean of 18.6 hours and a standard deviation of 6 hours. For simple random samples of size n=36 from this population, you determine that the sampling distribution of the sampling distribution of the sample mean is approximately normal with a mean of 18.6 hours and a standard deviation of 6/sqrt(36)=1 hour. If you discover that Northside High has only 200 seniors, which of the following would no longer be correct?
- the mean and the standard deviation
- the standard deviation and the shape of the distribution
- All three would be incorrect.
- only the standard deviation
- only the shape of the distribution

Transcript text: Suppose that at Northside High, the number of hours per week that seniors spend on homework is approximately Normally distributed with a mean of 18.6 hours and a standard deviation of 6 hours. For simple random samples of size $n=36$ from this population, you determine that the sampling distribution of the sampling distribution of the sample mean is approximately normal with a mean of 18.6 hours and a standard deviation of $\frac{6}{\sqrt{36}}=1$ hour. If you discover that Northside High has only 200 seniors, which of the following would no longer be correct? the mean and the standard deviation the standard deviation and the shape of the distribution All three would be incorrect. only the standard deviation only the shape of the distribution

Solution

Solution Steps

Step 1: Understand the given information

The population of seniors at Northside High has a mean ($\mu$) of 18.6 hours and a standard deviation ($\sigma$) of 6 hours. The sampling distribution of the sample mean for $n = 36$ is approximately normal with a mean of 18.6 hours and a standard deviation of $\frac{6}{\sqrt{36}} = 1$ hour. However, the population size is only 200 seniors.

Step 2: Check the conditions for the sampling distribution

For the sampling distribution of the sample mean to be approximately normal, the sample size $n$ should be large enough (typically $n \geq 30$) and the population size $N$ should be much larger than $n$. Here, $n = 36$ and $N = 200$. Since $n$ is not negligible compared to $N$ (specifically, $n > 0.05N$), the finite population correction factor (FPC) should be applied to the standard deviation of the sampling distribution.

The FPC is given by: \[ \text{FPC} = \sqrt{\frac{N - n}{N - 1}} \] Substituting $N = 200$ and $n = 36$: \[ \text{FPC} = \sqrt{\frac{200 - 36}{200 - 1}} = \sqrt{\frac{164}{199}} \approx 0.9074 \] Thus, the corrected standard deviation of the sampling distribution is: \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \cdot \text{FPC} = \frac{6}{\sqrt{36}} \cdot 0.9074 = 1 \cdot 0.9074 = 0.9074 \]

Step 3: Determine which statements are incorrect

The mean of the sampling distribution remains correct at 18.6 hours.
The standard deviation of the sampling distribution is no longer 1 hour; it is now approximately 0.9074 hours.
The shape of the sampling distribution is still approximately normal because $n = 36$ is large enough for the Central Limit Theorem to apply, even with the FPC.

Thus, only the standard deviation is incorrect.