Here are 6 measurements of the electrical conductivity of a iron rod:
10.08, 9.89,10.05,10.16,10.21,1011
The iron rod is supposed to have conductivity 10.1.
Do the measurements give good evidence that the true conductivity is not 10.1?
The 6 measurements are an SRS from the population of all results we would get if we kept measuring conductivity forever.
This population has a Normal distribution with mean equal to the true conductivity of the iron rod and standard deviation 0.1
jhvhn
test
To determine if the measurements provide evidence that the true conductivity is not 10.1, we can perform a hypothesis test.
Step 1: State the null and alternative hypotheses.
The null hypothesis (H0) is that the true conductivity is 10.1.
The alternative hypothesis (Ha) is that the true conductivity is not 10.1.
Step 2: Decide on a significance level.
Choose a significance level (often denoted as α) to determine the threshold at which we reject the null hypothesis. Let's assume a significance level of 0.05 (or 5%).
Step 3: Calculate the test statistic.
In this case, we can use the sample mean to calculate the test statistic. The formula for the test statistic is:
t = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the hypothesized true conductivity (10.1), s is the sample standard deviation, and n is the sample size.
Step 4: Determine the critical value(s).
To determine the critical value(s), we need to look up the value(s) from the t-distribution table based on the degrees of freedom (n-1) and the significance level. Since we have six measurements, the degrees of freedom is 5.
Step 5: Compare the test statistic with the critical value(s).
If the test statistic falls outside the critical value(s) range, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Step 6: Make a conclusion.
Based on the comparison in Step 5, we can decide whether to reject or fail to reject the null hypothesis. If we reject the null hypothesis, it means there is sufficient evidence that the true conductivity is not 10.1.
Please provide the sample standard deviation (s) and the sample size (n) to proceed with the calculation of the test statistic and the critical values.
To determine whether the measurements provide good evidence that the true conductivity is not 10.1, we can conduct a hypothesis test. Assuming that the population of all conductivity measurements follows a normal distribution with a mean equal to the true conductivity of the iron rod (10.1) and a standard deviation of 0.1, we can perform a one-sample t-test.
Let's outline the steps to conduct the hypothesis test:
Step 1: State the null and alternative hypotheses:
The null hypothesis (H0) states that the true conductivity of the iron rod is equal to 10.1.
The alternative hypothesis (H1) states that the true conductivity of the iron rod is not equal to 10.1.
Step 2: Set the significance level:
Choose a significance level (alpha) to determine the threshold for rejecting the null hypothesis. Common values for alpha include 0.05 or 0.01.
Step 3: Calculate the test statistic:
Calculate the test statistic using the formula:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
Step 4: Determine the critical region:
Based on the chosen significance level and the degrees of freedom (sample size - 1), find the critical t-value(s) from the t-distribution table. The critical region lies in the tails of the distribution.
Step 5: Compare the test statistic with critical values:
If the absolute value of the test statistic falls within the critical region, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Now, let's calculate the test statistic and perform the hypothesis test using the provided measurements:
Sample mean = (10.08 + 9.89 + 10.05 + 10.16 + 10.21 + 1011) / 6 = 1691.3 / 6 = 281.88
Sample standard deviation = sqrt(((10.08 - 281.88)^2 + (9.89 - 281.88)^2 + (10.05 - 281.88)^2 + (10.16 - 281.88)^2 + (10.21 - 281.88)^2 + (1011 - 281.88)^2) / (6 - 1)) = 501932.997
t = (281.88 - 10.1) / (501932.997 / sqrt(6))
Next, we need to find the critical t-value(s) based on the chosen significance level and degrees of freedom (5, since sample size - 1 = 6 - 1 = 5).
Lastly, we compare the calculated t-value with the critical t-value(s) to determine whether to reject the null hypothesis or fail to reject it.
Unfortunately, we do not have enough information to calculate the exact values or the resulting hypothesis test. The provided dataset seems to contain an outlier with a value of 1011, which might indicate a data entry error or an outlier for the measurements. To ensure reliable results, it's crucial to identify and address any outliers or errors before proceeding with the hypothesis test.