In other words, the sample provides strong evidence that the average number of cigarettes smoked per day is significantly lower than the reported average of 31 cigarettes.

To answer this question, we can perform a one-sample t-test with the following hypotheses.

Diketahui :

• t = (x̄ - μ) / (s / √n)

Ditanya :

• Is thare enought evidence to support her claim?

Jawab :

Penjelasan dengan langkah-langkah:

To answer this question, we can perform a one-sample t-test with the following hypotheses:

• Null hypothesis (H0): The true mean number of cigarettes smoked per day is 31.
• Alternative hypothesis (Ha): The true mean number of cigarettes smoked per day is less than 31.
• We will use a significance level (alpha) of 0.05.

The test statistic for the one-sample t-test is:

t = (x̄ - μ) / (s / √n)

where:

x̄ is the sample mean (28)

μ is the hypothesized population mean (31)

s is the sample standard deviation (2.7)

n is the sample size (10)

Plugging in the numbers, we get:

t = (28 - 31) / (2.7 / √10) = -3.68

Using a t-distribution table with 9 degrees of freedom (n-1), we find that the critical t-value for a one-tailed test with alpha=0.05 is -1.833.

Since our calculated t-value of -3.68 is less than the critical t-value of -1.833, we can reject the null hypothesis and conclude that there is enough evidence to support the researcher's claim that the true mean number of cigarettes smoked per day is less than 31. In other words, the sample provides strong evidence that the average number of cigarettes smoked per day is significantly lower than the reported average of 31 cigarettes.

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