A drone builder states in their terms of conditions that the battery for their foldable drone will last for more than 4 hours of continuous flight-time if the battery was fully charged before the flight. To test this, claim a sample of n = 10 drones were tested. The results showed a sample mean of 4.2 hours and a sample standard deviation of 0.4 hours. Conduct the hypothesis test using a 0.5 level of significance and determine whether or not the manufacturers claim is supported.

Answer:

The manufacturer's claim is not supported

Step-by-step explanation:

The time duration the battery of the foldable drone is said to last, μ = 4 hours

The number of drones that were tested, n = 10 drones

The sample mean, [tex]\overline x[/tex] = 4.2 hours

The standard deviation, s = 0.4 hours

The hypothesis test level of significance = 0.5

The null hypothesis, H₀; μ = 4

The alternative hypothesis, Hₐ; μ ≠ 4

The test statistic is given as follows;

[tex]t=\dfrac{\bar{x}-\mu }{\dfrac{s }{\sqrt{n}}}[/tex]

We get;

[tex]The \ test \ statistic, \ t =\dfrac{4.2 - 4}{\dfrac{0.4}{\sqrt{10} } } \approx 1.58113883008[/tex]

Therefore, the test statistic, t ≈ 1.58

The degrees of freedom, d. f. = n - 1

For n = 10, we have;

The degrees of freedom, d. f. = 10 - 1 = 9

From the t-table at 0.5 level of significance, the critical-t = 0.7

Given that the test statistic is larger than the critical-t, we reject the null hypothesis and there is enough statistical evidence to suggest that the manufacturer claim is not supported and that the mean is not 4 hours as claimed.

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