Question

Researchers studied the compression strength in lbs of 12-oz aluminum cans. A simple random sample of 35 aluminum cans filled with cola (which has more carbonation) had a sample mean compression strength of 554 lbs and a sample standard deviation of 15 lbs. A simple random sample of 30 aluminum cans filled with strawberry soda had a sample mean compression strength of 540 lbs and a sample standard deviation of 21 lbs. At the 0.05 significance level, perform a hypothesis test to determine if the mean compression strength, Hy, of aluminum cans filled with cola is more than the mean compression strength, My, of aluminum cans filled with strawberry soda. 1. Select the appropriate null and alternate hypotheses. OH, MM Hq Hq <h OH, My = H2, H, H2> OH, H4> H2, H, 14 =12 OH, ty > H2, H, 44<H2 OH, My = H2, H, MyM2 OHO My H₂ H₃ Hq FH2 2. Find the value of the test statistic. Round to three decimal places. 3. Find the P-value. Round to four decimal places. 4. Select the correct interpretation of the results. O At the 0.01 significance level, there is not enough evidence that the mean compression strength of aluminum cans filled with cola is more than the mean compression strength of aluminum cans filled with strawberry soda.

Answer 1

Solution :

1) Null and alternative hypotheses :

The null and alternative hypotheses would be as follows :

Ho: Mi = 42

$\large H_{1}: \mu_{1}>\mu_{2}$

2) Test statistic :

To test the hypothesis we shall use two independent samples t-test assuming equal population variances. The test statistic is given as follows :

(11 – T2) t= S2 ( n + b) pooled

Where, Spooled (n1 – 1)si + (n2 – 1)sı ni + n2 - 2

x̄_{​​​​​1}, x̄_{​​​​​2} are sample means, s_{​​​​​​1}, s_{​​​​​​2} are sample standard deviations and n_{​​​​​1}, n_{​​​​​​2} are sample sizes.

We are given the following values :

$\large \bar{x}_{1} = 554,\ \ \bar{x}_{2} = 540$

$\large s_{1} = 15,\ \ s_{2} = 21$

$\large n_{1} = 35,\ \ n_{2} = 30$

$\large S^{2}_{pooled} = \frac{(35-1)(15)^{2}+(30-1)(21)^{2}}{35+30-2} = 324.4286$

$\large \therefore t = \frac{(554-540)}{\sqrt{324.4286\times\left ( \frac{1}{35}+\frac{1}{30} \right ) }} =3.124$

The value of the test statistic is 3.124.

3) P-value :

Since, our test is right-tailed test, therefore we shall obtain right-tailed p-value for the test statistic. The right-tailed p-value is given as follows :

P-value = P(T > t)

P-value = P(T > 3.124)

P-value = 0.0013

The p-value is 0.0013.

Decision :

Significance level = 0.05

P-value = 0.0013

(0.0013 < 0.05)

Since, p-value is less than the significance level of 0.05, therefore we shall reject the null hypothesis (H_{​​​​​​0}) at 0.05 significance level.

4) Conclusion :

At 0.05 significance level, there is enough evidence that the mean compression strength of aluminum cans filled with cola is more than the mean compression strength of aluminum cans filled with strawberry soda.

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