The producers of a new toothpaste claim that it prevents more cavities than other brands of toothpaste. A random sample of 60 people use the new toothpaste for 6 months. The mean number of cavities at their next check up is 1.5. In the general population the mean # of cavities after a 6 month check up is 1.73 (@= 1.12).
a) is it 1 or 2 tailed test? (I think one- tailed)
b) What are the Ho and Ha for this study? I think:
Ho: µ = µ1 or µ cavities from new toothpaste= cavities any other brand toothpaste
Ho: µ0>µ1 or µ cavities from new toothpaste> µ cavities with any other brand toothpaste
Ha: µ0<µ1 or µ cavities from new toothpaste<µ cavities with any other brand toothpaste
c) Compute Zobt -I don't know how
d) What is Z cv?- Also don't know.
e) Should Ho be rejected (I think yes)
f) Determine the 95% confidence interval for the population mean, based on the sample mean.
Thank you in advance for your help!
I would phrase the hypotheses this way:
Ho: µ ≥ 1.73
Ha: µ < 1.73
This would be a one-tailed test since Ha is showing a specific direction (less than).
Use a z-test:
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
Once you have the z-test statistic, compare to a critical value from a z-table. If the test statistic exceeds the critical value from the table, reject the null. If the test statistic does not exceed the critical value from the table, you cannot reject the null. You can draw your conclusions from there.
From looking at the data, it does appear Ho will be rejected.
Confidence interval formula:
CI95 = mean ± 1.96(sd/√n)
...where sd = standard deviation and n = sample size.
I hope these few hints will help get you started.
a) To determine whether it is a one-tailed or two-tailed test, we need to examine the alternative hypothesis (Ha). In this case, the alternative hypothesis states that the mean number of cavities for the new toothpaste is less than the mean number of cavities for other brands. Since the alternative hypothesis is one-sided (less than), it is a one-tailed test.
b) The null hypothesis (Ho) is a statement of no difference or no effect. In this case:
Ho: µ new toothpaste = µ other brands toothpaste
The alternative hypothesis (Ha) is a statement of the direction or effect we want to prove. In this case:
Ha: µ new toothpaste < µ other brands toothpaste
c) To compute Zobt (observed Z-value), we need to use the following formula:
Zobt = (x̄ - µ) / (σ / √n)
Where:
x̄ is the sample mean (1.5)
µ is the population mean (1.73)
σ is the population standard deviation (1.12)
n is the sample size (60)
Substituting the given values:
Zobt = (1.5 - 1.73) / (1.12 / √60)
d) To determine Zcv (critical value), we need to consider the significance level, which is not mentioned in the question. Let's assume a significance level of 0.05 (or 95% confidence).
For a one-tailed test with a significance level of 0.05, the critical value corresponds to a Z-value at which 95% of the area under the standard normal curve lies in the left tail (since the alternative hypothesis is µ new toothpaste < µ other brands toothpaste).
Using a standard normal distribution table or calculator, the Zcv for a significance level of 0.05 is approximately -1.645.
e) To determine whether Ho should be rejected, we compare the observed Z-value (Zobt) to the critical value (Zcv). If Zobt is less than Zcv, we reject Ho; otherwise, we fail to reject Ho.
If Zobt < Zcv, then we reject Ho and conclude that the new toothpaste has a statistically significant effect in preventing cavities compared to other brands.
f) To determine the 95% confidence interval for the population mean, we can use the sample mean (x̄), the sample size (n), and the standard deviation (σ).
The formula for the confidence interval is:
CI = x̄ ± (Z * (σ / √n))
Using the given values, we can calculate the confidence interval. Since no values for σ are provided, we can estimate it using the sample standard deviation (s) if it is available. If not, we will have to assume σ = 1.12 (population standard deviation).
Please provide the value of s (sample standard deviation) or let me know if you want to assume σ = 1.12 for the calculation of the confidence interval.
a) It is a one-tailed test.
b) The null hypothesis (Ho) is that the mean number of cavities using the new toothpaste is equal to the mean number of cavities using any other brand of toothpaste (µ0 = µ1).
The alternative hypothesis (Ha) is that the mean number of cavities using the new toothpaste is less than the mean number of cavities using any other brand of toothpaste (µ0 < µ1).
c) To compute Zobt (observed Z-value), we need to use the formula:
Zobt = (x̄ - µ0) / (σ/√n)
Where:
x̄ is the sample mean (1.5),
µ0 is the null hypothesis mean (1.73),
σ is the population standard deviation (1.12),
n is the sample size (60).
Substituting the values into the equation:
Zobt = (1.5 - 1.73) / (1.12/√60)
Calculating further:
Zobt = (-0.23) / (1.12/√60)
= (-0.23) / (1.12/7.746)
≈ -1.680
Therefore, Zobt is approximately -1.680.
d) The Zcv (critical value of Z) is the value associated with the chosen significance level and direction of the test. Since the alternative hypothesis is one-tailed, and assuming a significance level of α = 0.05 (95% confidence level), we need to find the Z-value that corresponds to an area of 0.05 in the tail of the standard normal distribution.
Using a Z-table or calculator, the Zcv for a one-tailed test with a significance level of α = 0.05 is approximately -1.645 (rounding to three decimal places).
e) To determine if the null hypothesis should be rejected, compare the Zobt to the Zcv.
If Zobt < Zcv, then we reject the null hypothesis (Ho). In this case, Zobt (-1.680) < Zcv (-1.645), which means we can reject the null hypothesis.
f) To determine the 95% confidence interval for the population mean based on the sample mean, we can use the formula:
Confidence Interval = x̄ ± Zcv * (σ/√n)
Where x̄ is the sample mean (1.5),
Zcv is the Z-value for the desired confidence level,
σ is the population standard deviation (1.12),
n is the sample size (60).
Substituting the values into the equation:
Confidence Interval = 1.5 ± (-1.645) * (1.12/√60)
Calculating further:
Confidence Interval = 1.5 ± (-1.645) * (1.12/7.746)
= 1.5 ± (-0.239)
Therefore, the 95% confidence interval for the population mean is approximately (1.261, 1.739).