A hospital director is told that $55 \%$ of the treated patients are insured. The director wants to test the claim that the percentage of insured patients is under the expected percentage. A sample of 300 patients found that 150 were insured. Determine the $P$-value of the test statistic. Round your aniswer to four decimal places

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Step 1 ：The problem is asking for the P-value of the test statistic. The P-value is the probability that a random chance generated the data, or something else that is equal or more extreme. In this case, we are testing the claim that the percentage of insured patients is under the expected percentage.

Step 2 ：We know that the expected proportion of insured patients is 0.55, and we have a sample of 300 patients with 150 insured. This gives us a sample proportion of 150/300 = 0.5.

Step 3 ：We can use the formula for the test statistic in a one-sample proportion hypothesis test, which is \((p\_hat - p) / \sqrt{(p*(1-p))/n}\), where \(p\_hat\) is the sample proportion, \(p\) is the expected proportion, and \(n\) is the sample size.

Step 4 ：We can then use the normal distribution to find the P-value. The P-value is the probability that a standard normal random variable is less than the test statistic.

Step 5 ：Let's calculate the test statistic and the P-value. \(p = 0.55\), \(p\_hat = 0.5\), \(n = 300\), the test statistic is -1.7407765595569797, and the P-value is 0.0409.

Step 6 ：Final Answer: The $P$-value of the test statistic is \(\boxed{0.0409}\).