Question

A researcher would like to determine whether an over-the-counter herbal nasal spray has an effect on mental alertness for infants. A sample of n = 18 infants is obtained, and each baby is given a standard dose of the spray one hour before being tested for mental alertness. For the general population of infants, scores on a mental alertness screening test are normally distributed with μ = 60. The infants in the sample had an average score of M = 54 and an s = 8.

Can the researcher conclude that scores on the screening test are significantly different after administering the spray? Use a two-tailed test with α = .05. Be sure to state your null hypothesis, a decision about that null hypothesis, and a conclusion explaining your results with an APA format statement. Calculate Cohen’s d if necessary.

Answer 1

The provided sample mean is Xˉ=54

and the sample standard deviation is s = 8,

and the sample size is n = 18.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 60

Ha: μ ≠ 60

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05,

and the critical value for a two-tailed test is

t_c = 2.11

The rejection region for this two-tailed test is

R={t:∣t∣>2.11}

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that

∣t∣=3.182>tc​=2.11,

it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p = 0.0055,

and since p = 0.0055<0.05,

it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected.

Therefore, there is enough evidence to claim that the population mean μ is different than 60, at the 0.05 significance level.

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