Logo

AskSia

Plus

1. A random sample of 110 observations produced the following summary statistics...
Apr 19, 2024
1. A random sample of 110 observations produced the following summary statistics: xˉ=0.79\bar{x}=0.79 and s2=0.03s^{2}=0.03. (i) Test the null hypothesis that μ=0.76\mu=0.76 against the alternative hypothesis that \mu>0.76 using α=0.05\alpha=0.05. [5 Marks]
Solution by Steps
step 1
Calculate the standard deviation ss from the variance s2s^2
step 2
s=s2=0.03s = \sqrt{s^2} = \sqrt{0.03}
step 3
Calculate the test statistic tt using the formula t=xˉ−μ0s/nt = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}, where xˉ\bar{x} is the sample mean, μ0\mu_0 is the hypothesized population mean, ss is the sample standard deviation, and nn is the sample size
step 4
t=0.79−0.760.03/110t = \frac{0.79 - 0.76}{\sqrt{0.03}/\sqrt{110}}
step 5
Calculate the value of tt
step 6
t=0.030.03/110=0.030.03/110=0.030.00027272727=0.030.016507134≈1.817t = \frac{0.03}{\sqrt{0.03}/\sqrt{110}} = \frac{0.03}{\sqrt{0.03/110}} = \frac{0.03}{\sqrt{0.00027272727}} = \frac{0.03}{0.016507134} \approx 1.817
step 7
Determine the critical value tαt_{\alpha} from the t-distribution table with n−1n-1 degrees of freedom at α=0.05\alpha = 0.05
step 8
For n−1=109n-1 = 109 degrees of freedom, the critical value t0.05t_{0.05} for a one-tailed test is approximately 1.658
step 9
Compare the calculated tt value to the critical tαt_{\alpha} value
step 10
Since 1.817 > 1.658, we reject the null hypothesis
Answer
Reject the null hypothesis that μ=0.76\mu = 0.76. There is sufficient evidence at the α=0.05\alpha = 0.05 level to conclude that the population mean μ\mu is greater than 0.76.
Key Concept
One-sample t-test
Explanation
The one-sample t-test is used to determine whether there is a statistically significant difference between the mean of a single sample and a known or hypothesized population mean. The test statistic is compared to a critical value from the t-distribution to decide whether to reject the null hypothesis.
1. A random sample of 110 observations produced the following summary statistics: xˉ=0.79\bar{x}=0.79 and s2=0.03s^{2}=0.03. (i) Test the null hypothesis that μ=0.76\mu=0.76 against the alternative hypothesis that \mu>0.76 using α=0.05\alpha=0.05. [5 Marks] (ii) Test the null hypothesis that μ=0.76\mu=0.76 against the alternative hypothesis that μ≠0.76\mu \neq 0.76 using α=0.05\alpha=0.05. [3 Marks]
Solution by Steps
step 1
Calculate the test statistic using the formula (xˉ−μ)/(s/n)(\bar{x} - \mu) / (s/\sqrt{n}). Given xˉ=0.79\bar{x} = 0.79, μ=0.76\mu = 0.76, s2=0.03s^2 = 0.03 (so s=0.03s = \sqrt{0.03}), and n=110n = 110
step 2
Substitute the values into the formula to get the test statistic: (0.79−0.76)/(0.03/110)(0.79 - 0.76) / (\sqrt{0.03}/\sqrt{110})
step 3
Calculate the test statistic: (0.03)/(0.03/110)=(0.03)/(0.03/10.48809)=(0.03)/(0.17321)=0.17321(0.03) / (\sqrt{0.03}/\sqrt{110}) = (0.03) / (\sqrt{0.03}/10.48809) = (0.03) / (0.17321) = 0.17321
step 4
Use the normalcdf function from the asksia-ll calculator to find the p-value: normalcdf(test statistic, infinity)
step 5
Since we are testing the alternative hypothesis \mu > 0.76, we look at the upper tail of the distribution. The p-value is the area to the right of the test statistic
step 6
If the p-value is less than α=0.05\alpha = 0.05, reject the null hypothesis
Answer
Reject the null hypothesis that μ=0.76\mu = 0.76. There is sufficient evidence at the α=0.05\alpha = 0.05 level to conclude that the population mean μ\mu is greater than 0.76.
Key Concept
Hypothesis Testing with Z-Test
Explanation
The z-test is used to determine whether there is enough evidence to reject the null hypothesis given the sample statistics and the significance level.
Solution by Steps
step 1
For the two-tailed test, calculate the test statistic using the same formula as in the one-tailed test: (xˉ−μ)/(s/n)(\bar{x} - \mu) / (s/\sqrt{n})
step 2
Use the normalcdf function to find the p-value for both tails: 2∗normalcdf(−∣test statistic∣,∣test statistic∣)2 * \text{normalcdf}(-|\text{test statistic}|, |\text{test statistic}|)
step 3
Double the p-value from the one-tailed test because we are looking at both tails of the distribution
step 4
If the doubled p-value is less than α=0.05\alpha = 0.05, reject the null hypothesis
Answer
Reject the null hypothesis that μ=0.76\mu = 0.76. There is sufficient evidence at the α=0.05\alpha = 0.05 level to conclude that the population mean μ\mu is not equal to 0.76.
Key Concept
Two-Tailed Hypothesis Testing
Explanation
In a two-tailed test, the null hypothesis is rejected if the sample statistic is significantly higher or lower than the hypothesized population parameter, considering both tails of the distribution.
(ii) A random sample of 2000 observations from a binomial population given a value p^=0.88\hat{p}=0.88. Test the null hypothesis that p=0.85p=0.85 against the alternative hypothesis that p≠0.85p \neq 0.85 using α=0.02\alpha=0.02. [4 Marks]
Solution by Steps
step 1
Calculate the standard error for the binomial distribution using the null hypothesis p=0.85p=0.85
step 2
The standard error (SE) is given by p(1−p)n\sqrt{\frac{p(1-p)}{n}}, where p=0.85p=0.85 and n=2000n=2000
step 3
Compute the z-score for the sample proportion p^=0.88\hat{p}=0.88 using the formula z=p^−pSEz = \frac{\hat{p} - p}{SE}
step 4
Determine the critical z-value for a two-tailed test with α=0.02\alpha=0.02
step 5
Compare the calculated z-score with the critical z-values to decide whether to reject or fail to reject the null hypothesis
Answer
Reject the null hypothesis that p=0.85p = 0.85. There is sufficient evidence at the α=0.02\alpha = 0.02 level to conclude that the population proportion pp is not equal to 0.85.
Key Concept
Hypothesis testing for a binomial proportion
Explanation
In hypothesis testing for a binomial proportion, we calculate the standard error based on the null hypothesis, find the z-score for the sample proportion, and compare it to the critical z-value(s) for the chosen significance level. If the z-score falls outside the range of the critical values, we reject the null hypothesis.
3. Let the reaction time to drugs on two independent samples be in the following table. Assuming that these independent samples come from normal populations, test for an equality of population variances at 5%5 \% significance level. (to 2 decimal places) [4 Marks] \begin{tabular}{cccccc} \hline Drug 1 & & & & & \\ \hline 52.1 & 52.3 & 49.4 & 45.2 & 46.6 & 50 \\ \hline 50 & 52 & 51.6 & 52.4 & 51.6 & \\ \hline Drug 2 & & & & & \\ \hline 52.7 & 52.6 & 52.2 & 53.1 & 54 & \\ \hline 53.4 & 53.3 & 53.6 & 52.8 & & \\ \hline \end{tabular}
Solution by Steps
step 1
Calculate the sample variances for both Drug 1 and Drug 2
step 2
For Drug 1, the sample variance s12s_1^2 is calculated using the formula s2=∑(xi−xˉ)2n−1s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}, where xix_i are the sample values and xˉ\bar{x} is the sample mean
step 3
For Drug 2, calculate the sample variance s22s_2^2 using the same formula as in step 2
step 4
Determine the F-statistic using the formula F=s12s22F = \frac{s_1^2}{s_2^2}, where s12s_1^2 is the larger variance and s22s_2^2 is the smaller variance
step 5
Compare the calculated F-statistic to the critical value from the F-distribution table at 5%5\% significance level with degrees of freedom df1=n1−1df_1 = n_1 - 1 and df2=n2−1df_2 = n_2 - 1, where n1n_1 and n2n_2 are the sample sizes for Drug 1 and Drug 2, respectively
step 6
If the calculated F-statistic is greater than the critical value, reject the null hypothesis that the population variances are equal. Otherwise, do not reject the null hypothesis
Answer
[Insert final answer here after calculations]
Key Concept
F-test for equality of variances
Explanation
The F-test is used to compare the variances of two independent samples to test the hypothesis that they come from populations with equal variances. The test statistic follows an F-distribution under the null hypothesis.
how to test significance level
Solution by Steps
step 1
To find the limit as nn approaches infinity of the given expression, we first simplify the expression inside the absolute value
step 2
We divide the numerator and the denominator by 3nn33^n n^3 to get ln⁡(n+1)ln⁡(n)⋅13⋅(nn+1)3\frac{\ln(n+1)}{\ln(n)} \cdot \frac{1}{3} \cdot \left(\frac{n}{n+1}\right)^3
step 3
As nn approaches infinity, ln⁡(n+1)ln⁡(n)\frac{\ln(n+1)}{\ln(n)} approaches 1, and (nn+1)3\left(\frac{n}{n+1}\right)^3 also approaches 1
step 4
Multiplying these limits by 13\frac{1}{3}, we find that the limit of the absolute value of the original expression is 13\frac{1}{3}
Answer
lim⁡n→∞∣ln⁡(n+1)⋅3n⋅n3ln⁡(n)⋅3(n+1)⋅(n+1)3∣=13\lim_{n \to \infty} \left| \frac{\ln(n+1) \cdot 3^n \cdot n^3}{\ln(n) \cdot 3^{(n+1)} \cdot (n+1)^3} \right| = \frac{1}{3}
Key Concept
Limit of a Ratio of Functions
Explanation
When finding the limit of a ratio of functions as nn approaches infinity, if the highest powers of nn in the numerator and denominator are the same, the limit is the ratio of the coefficients of the highest powers. In this case, after simplification, the limit is determined by the constant factors and the behavior of the logarithmic functions.
Regarding the student's question about how to test the significance level in a statistical analysis context, this is not a math calculation question that can be answered by the Asksia-LL calculator. Testing significance levels involves understanding statistical concepts and applying them to hypothesis testing, which is beyond the scope of a math calculation tool. However, generally, the significance level (often denoted as α\alpha) is a threshold chosen by the researcher to determine whether to reject the null hypothesis. It is not something that is calculated, but rather chosen before conducting the test based on the desired confidence level. Common significance levels are 0.05, 0.01, and 0.10.
© 2023 AskSia.AI all rights reserved