researchers believe a new sleep on average 10 hours per night ( µ = 10), with a population standard deviation of 2.3 hours(s=2.3) it is assume that the number of hours of sleep per night is normal distributed ,

however you think this drug causes people to sleep less than 10 hours per night. you select of 25 people to test this claim. therefore the hypotheses are Ho: µ=140 He:µ<10 the test reject Ho at roughly if x < 9.24 hours per night . find the probability of a type 1 error then find the probability of type 2 error when on fact µ=8.2 hours of sleep per night .

To find the probability of a Type 1 error, we need to calculate the significance level (α) of the test. In this case, the test rejects the null hypothesis (Ho: µ = 10) when the observed sample mean (x̄) is less than 9.24 hours per night.

Since the test is one-tailed (rejecting if x < 9.24), we can find the probability of Type 1 error by calculating the area under the null hypothesis distribution curve to the left of 9.24.

To do this, we need to standardize the test statistic using the population standard deviation and the sample size:

z = (x̄ - µ) / (σ / sqrt(n))

where x̄ is the sample mean, µ is the hypothesized population mean, σ is the population standard deviation, and n is the sample size.

In this case, x̄ = 9.24, µ = 10, σ = 2.3, and n = 25.

Substituting these values into the formula:

z = (9.24 - 10) / (2.3 / √25)
z = -0.76

Using a standard normal distribution table or calculator, we can find the probability of a Type 1 error by finding the area to the left of -0.76.

The probability of a Type 1 error, denoted as α, is the significance level of the test. Let's assume a significance level of 0.05 (5%).

By consulting a standard normal distribution table or using a calculator, we find that the area to the left of -0.76 is approximately 0.2236. Therefore, the probability of a Type 1 error is approximately 0.2236.

Now, let's calculate the probability of a Type 2 error. To do this, we need the specific alternative hypothesis value for the population mean, which is µ = 8.2 hours of sleep per night.

We can calculate the test statistic for the alternative hypothesis:

z = (x̄ - µ) / (σ / sqrt(n))
z = (9.24 - 8.2) / (2.3 / √25)
z = 1.04

To find the probability of a Type 2 error, we will need to calculate the area under the null hypothesis distribution curve to the right of 1.04. This is the probability of failing to reject the null hypothesis when the alternative hypothesis is true.

Again, using a standard normal distribution table or calculator, we find that the area to the right of 1.04 is approximately 0.1492. Therefore, the probability of a Type 2 error is approximately 0.1492.

To find the probability of a Type 1 error, we need to assume that the null hypothesis (Ho) is true, i.e., µ = 10 hours. In this case, a Type 1 error occurs when we incorrectly reject the null hypothesis.

Given that the test rejects Ho at roughly x < 9.24 hours per night, we need to find the probability of observing a sample mean (x̄) less than 9.24 under the assumption that µ = 10.

To calculate this probability, we can use the z-score formula:

z = (x̄ - µ) / (σ / √n)

In this case, x̄ = 9.24, µ = 10, σ = 2.3, and n = 25. Substituting these values:

z = (9.24 - 10) / (2.3 / √25)
= -0.76

Using a standard normal distribution table or a calculator, we can find the probability associated with a z-score of -0.76. This gives us the probability of observing a sample mean less than 9.24 under the assumption that µ = 10.

Now, to find the probability of a Type 2 error, we need to assume that the alternative hypothesis (He) is true, i.e., µ < 10 hours. In this case, a Type 2 error occurs when we fail to reject the null hypothesis when it is false.

Given that µ = 8.2 hours, we need to find the probability of observing a sample mean greater than or equal to 9.24 (the threshold for rejecting the null hypothesis). This represents the probability of failing to reject the null hypothesis when in fact µ = 8.2.

Using the same z-score formula:

z = (x̄ - µ) / (σ / √n)

Here, x̄ = 9.24, µ = 8.2, σ = 2.3, and n = 25. Substituting these values:

z = (9.24 - 8.2) / (2.3 / √25)
= 1.74

Again, using a standard normal distribution table or a calculator, we can find the probability associated with a z-score of 1.74. This gives us the probability of observing a sample mean greater than or equal to 9.24 under the assumption that µ = 8.2.

Please note that without specific probabilities associated with the null and alternative hypotheses, we cannot determine the exact probabilities of Type 1 and Type 2 errors. The calculations provided assume a standard normal distribution and serve as an example of how to calculate these probabilities.