Given The Sample Information, Test The Hypothesis ✓ Solved
Given the following sample information, test the hypothesis
Given the following sample information, test the hypothesis that the treatment means are equal at the .05 significance level. Treatment 1 Treatment 2 Treatment.
b) What is the decision rule? (Round your answer to 2 decimal places.) Reject H o if F >.
Compute SST, SSE, and SS total. (Round your answers to 2 decimal places.) SST SSE SS total.
(d) Complete an ANOVA table. (Round F, SS to 2 decimal places and MS to 3 decimal places.) Source SS df MS F Treatments Error Total.
4. Suppose the manufacturer of Advil, a common headache remedy, recently developed a new formulation of the drug that is claimed to be more effective. To evaluate the new drug, a sample of 270 current users is asked to try it. After a one-month trial, 251 indicated the new drug was more effective in relieving a headache. At the same time a sample of 390 current Advil users is given the current drug but told it is the new formulation. From this group, 352 said it was an improvement.
(1) State the decision rule for .01 significance level: H 0: π n ≤ π c; H 1: π n > π c. (Round your answer to 2 decimal places.) Reject H 0 if z >.
(2) Compute the value of the test statistic. (Do not round the intermediate value. Round your answer to 2 decimal places.) Value of the test statistic.
(3) Can we conclude that the new drug is more effective? Use the .01 significance level. H 0. We conclude that the new drug is more effective. Do not rejected cannot.
Paper For Above Instructions
This paper aims to perform hypothesis testing using the provided sample information to determine if the treatment means are statistically equal. The analysis will include the calculation of SST (Total Sum of Squares), SSE (Error Sum of Squares), and the completion of an ANOVA (Analysis of Variance) table. Moreover, we will review a case study involving a new formulation of Advil, analyzing user effectiveness perceptions.
1. Hypothesis Testing for Treatment Means
The null hypothesis (H0) states that there is no significant difference between the treatment means, while the alternative hypothesis (H1) states that at least one treatment mean is different. We will conduct the test using a significance level of α = 0.05.
1.1 Decision Rule
To set the decision rule, we must compute the critical value from the F-distribution for our ANOVA test. The critical value F at α = 0.05 will be obtained using the degrees of freedom for treatment (k-1) and error (N-k), where k is the number of treatments and N is the total number of observations. A decision rule would state: "Reject H0 if F > F-critical value."
1.2 Computing SST, SSE, and SS Total
Next, we need to compute various sums of squares which are essential for calculating the F-statistic. The formulas are:
- SST = Σ(n_i)*(X̄_i - X̄)^2, where n_i is the sample size for each treatment, X̄_i is the mean for each treatment, and X̄ is the overall mean.
- SSE = Σ(n_i)*(S_i^2), where S_i^2 is the variance for each treatment.
- SS Total = SST + SSE.
1.3 ANOVA Table
The ANOVA table summarizes the variability for the treatments and error. The table will provide values for SS (Sum of Squares), df (degrees of freedom), MS (Mean Square), and F (F-statistic).
| Source | SS | df | MS | F |
|---|---|---|---|---|
| Treatments | SST | k - 1 | SST/(k - 1) | MS Treatments / MS Error |
| Error | SSE | N - k | SSE/(N - k) | |
| Total | SS Total | N - 1 |
2. Effectiveness of New Advil Formulation
In the case study concerning Advil, the null hypothesis is H0: πn ≤ πc (the new drug is not more effective) against H1: πn > πc (the new drug is more effective). The significance level is set at α = 0.01.
2.1 Decision Rule
The decision rule states: "Reject H0 if z > z-critical value for α = 0.01." For a significance level of 0.01, the critical z-value can be found in z-tables and is approximately 2.33.
2.2 Test Statistic Calculation
The test statistic can be computed using the formula for proportion:
z = (p̂n - p̂c) / √(p̂(1-p̂)(1/n + 1/c))
where p̂n is the sample proportion of users who prefer the new formulation, and p̂c is that of current users. After computing this, round the value to two decimal places.
2.3 Conclusion
After calculating the test statistic, we compare our value with the critical z-value. If the calculated z is greater than 2.33, we reject H0, concluding that the new drug formulation is more effective at a 0.01 significance level. Conversely, if it is less, we fail to reject H0.
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