The Toylot company makes an electric train with a motor that it claims will draw an average of only 0.8 ampere (A) under a normal load. A sample of twelve motors was tested, and it was found that the mean current was x = 1.26 A, with a sample standard deviation of s = 0.40 A. Do the data indicate that the Toylot claim of 0.8 A is too low? (Use a 1% level of significance.
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
To determine if the Toylot claim of 0.8 A is too low, we can conduct a hypothesis test using the given data.
Step 1: Formulate the null and alternative hypotheses:
Null hypothesis (H0): The average current draw of the electric train's motor is 0.8 A.
Alternative hypothesis (H1): The average current draw of the electric train's motor is greater than 0.8 A.
Step 2: Determine the test statistic and the rejection region:
Since we are comparing the sample mean to a claimed population mean, we can use a one-sample t-test. Since we want to test if the mean is greater than the claim, we will use a one-tailed test.
The test statistic for a one-sample t-test is given by: t = (x - μ) / (s / √n),
where x is the sample mean, μ is the claimed population mean, s is the sample standard deviation, and n is the sample size.
The rejection region is determined by the significance level (α) and the degrees of freedom (df = n - 1). In this case, the significance level is 1% (0.01) and the sample size is 12, so the degrees of freedom is 11. We will compare the calculated t-value to the critical t-value at α = 0.01 with df = 11.
Step 3: Calculate the test statistic:
Given:
x = 1.26 A (sample mean)
μ = 0.8 A (claimed population mean)
s = 0.40 A (sample standard deviation)
n = 12 (sample size)
t = (1.26 - 0.8) / (0.40 / √12)
= 0.46 / (0.40 / 3.464)
= 1.1526
Step 4: Compare the test statistic to the critical value:
Based on the given significance level of 1%, we need to find the critical t-value. With df = 11, the critical t-value for a one-tailed test at α = 0.01 is approximately 2.718.
Since our calculated t-value (1.1526) does not exceed the critical t-value (2.718), we fail to reject the null hypothesis.
Step 5: Make a conclusion:
Since we failed to reject the null hypothesis, we do not have enough evidence to conclude that the Toylot claim of 0.8 A is too low. The data does not indicate a significant difference between the claimed average current draw and the sample mean.
Note: It's important to note that failing to reject the null hypothesis does not mean the claim is definitely true. It simply means we do not have enough evidence to support an alternative hypothesis.